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MANUAL 


OF 


TRIGONOMETRY. 


FOR  THE  USE 


OF 


HIGH  SCHOOLS.  ACADEMIES  AND  COLLEGES. 


BY 

J.  B.  CLARKE,  Ph.  B. 

ASSISTANT   PROFESSOR  OF  MATHEMATICS 

IN   THE 

UNIVERSITY    OF    CALIFORNIA. 


\%V<. 


[AUTHOa'S    EDITION.] 


CC)?^H\Q»UT,  \%?.a,   QN   iOHH   B.   C\_NHY.£. 


F»ACIF^IC    PRESS, 

PRINTERS,   ELECTROTYPERS,  AND  BINDERS. 


PREFACE. 


In  the  following  pages  the  author  has  endeavored  to 
present  a  brief,  but  thorough  and  complete,  course  in 
Trigonometry.  While  non-essentials  have  been  excluded, 
considerable  attention  has  been  given  to  some  matters 
not  discussed  in  any  of  the  usual  text-books.  It  is  dis- 
couraging even  to  a  bright  student,  to  meet,  in  his  work 
in  the  Integral  Calculus,  difficulties  that  should  have 
been  surmounted  in  his  Trigonometrical  course.  The 
chapter  on  Algebraic  Trigonometry  will,  it  is  hoped, 
lighten  the  labor  of  the  student  preparing  for  advanced 
work  and  broaden  the  views  of  the  reader  whose  Trigo- 
nometry is  merely  a  minor  part  of  a  symmetrical  course 
of  liberal  ytudy.  The  text,  taken  in  connection  with  the 
exercises,  (all  of  which  should  be  worked  out),  will  afford 
ample  preparation  for  any  University  Mathematical 
course. 

The  author  desires  to  express  his  sense  of  obligation 
to  Prof.  Irving  Stringham  for  encouraging  and  valuable 
hints  and  suggestions. 


J.  B.  CLARKE. 


Department  of  Mathematics, 
University  of  California, 
February,  1888. 


183630 


TABLH  OK  CONTKNTS. 


Paoe. 

Chapter  I. — Preliminary  Definitions. — Signs  and  Limiting  Values 

of  Functions. — Fundamental  Relations. — Geometric  Repre- 
sentation of  Functions. — Trigonometric  Tables 9 

Chapter  II.  —Solution  of  Right  Triangles. — Solution  of  Oblique 

Triangles  by  means  of  Right  Triangles. — Examples 20 

Chapter  III. — Functions  of  the  Sum  and  of  the  Difference. — 
Sums  and  Differences  of  Functions. — Miscellaneous  Form- 
ulae        26 

Chapter  IV. — Additional  Formulae  for  Plane  Triangles. — Special 

Formulae 30 

Chapter  V. — Spherical  Trigonometey. — Napier's  Rules. — De- 
termination of  Species. — Solution  of  Right  Triangles  and 
Solution  of  Oblique  Triangles  by  means  of  Right  Triangles . .       33 

Section  II. — Additional  Formulae  for  Oblique  Triangles. — Special 

Formulae, — Napier's  and  Delambre's  Analogies  41 

Chapter  VI. — Algebraic  Trigonometry. — Trigonometric  De- 
velopments.— Exponential  Functions. — Circular  and  Hyper- 
bolic Functions. — Properties  and  Relations  of  the  Functions. — 
Geometric  Representation  of  the  Exponential  Functions 47 

Exercises. — 


OF  THE  \ 

IVERSfTT  ^ 

.     OF 


TRIGONOMETRY. 


CHAPTER   I. 


1.  Trigonometry  is  a  branch  of  Mathematics,  in  which 
are  discussed  the  properties,  relations,  and  applications  of  cer- 
tain functions  of  angles.  These  functions,  to  be  hereafter 
defined,  are  called  the  Trigonometric  functions,  and  are  practi- 
cally useful  mainly  in  the  solution  of  triangles. 

2.  An  angle,  AOB  (Fig.  1), 
is  considered  as  generated  by 
the  revolution  of  a  line  about 
the  vertex,  O,  from  the  posi- 
tion OA  to  the  position  OB. 
Ois  called  the  Angular  Centre; 
OA,  the  Initial,  and  OB,  the 
Tei-minal  dine.  We  call  the 
revolution,  and  the  angle  gen- 
erated, joo^i^ive,  when  the  motion 
is  opposite  in  direction  to  that  of 
fig- 1-  a   watch-hand ;   negative,  when 

the  motion  is  in  the  direction  of  that  of  a  watch-hand.  Angles 
are  assumed  to  be  positive  unless  they  are  explicitly  stated  to 
be  negative.  Angular  magnitude,  thus  considered,  evidently 
admits  of  infinite  extension ;  AOD  may  be  generated  by  a 
simple  passage  of  the  moving  line  from  OA  to  OD,  or  by  a 
movement  in  which  this  line  reaches  OD  after  passing  around 
O  any  number  of  times.  This  requires,  of  course,  that  the 
magnitude  of  the  measuring  arcs  be  similarly  considered ; 
hence  the  occurrence  of  angles  of  540°,  630°,  etc.   (i.    e.,  the 

(9) 


HHkSBSUHE 


10  DEFINITIONS. 

angle  as  ordinarily  taken,  +  any  multiple  of  360°).  The 
plane  space  about  O  is,  for  convenience,  divided  into  four  equal 
parts,  called  quadrants,  by  two  lines,  viz.,  the  Initial  line,  as- 
sumed horizontal  and  unlimited  in  length,  and  a  perpendicular 
thereto  through  O.  These  quadrants  are  numbered  as  in  the 
figure,  and  an  angle  lies  in  that  quadrant  in  which  is  found  its 
terminal  line. 

3.  An  angle  is  usually  denoted  by  a  single  symbol,  thus :  x, 
y,  A.  In  general  this  symbol  will  denote  the  radial  measure 
of  the  angle,  except  when  we  deal  with  triangles.  The  radial 
measure  (Ven.  IV,  19)  of  an  angle  is  the  arc  intercepted  (by 
the  initial  and  terminal  lines)  divided  by  the  radius  with  which 
it  is  described  from  the  vertex  as  a  centre.  The  unit  is  the 
angle  whose  sides  intercept  on  the  circumference  an  arc  equal 
in  length  to  the  radius,  for,  in  the  case  of  this  angle,  arc :  ra- 
dius =  1.  As  2-R  =  360°,  R  =  360  ^  2r  =  180°  -r  3.1415926  = 
57°.29578,  or  about  57°.3  (57°  17'  45'^  nearly).  To  pass 
from  the  "  degree  "  measure  of  an  angle  to  its  "  radial "  meas- 
ure, we  therefore  divide  by  57°.3,  and  conversely. 

The  complement  of  an  angle  (or  arc)  is  a  right  angle  (or 
90°),  minus  the  angle  (or  arc);  the  supplement  is  the  difference 
between  the  angle  (or  arc)  and  two  right  angles  (180°). 

4.  Angles  are  dealt  with  by  means  of  certain  Trigonometric 
Functions.  If  a  perpendicular  be  dropped  from  any  point  of 
the  Terminal  line  on  the  (unlimited)  Initial  line,  a  right  triangle 
is  formed  called  the  Triangle  of  Reference.  The  following  ratios, 
involving  its  sides,  are  the  Trigonometric  Functions  of  the  angle. 

(1.)  The  perpendicular  (p)  divided  by  the  hypothenuse,  is 
the  SINE   (sin).     Thus   (Fig.   1),  BM:OB  =  sin  AOB. 

(2.)  The  base  (6)  divided  by  the  hypothenuse,  is  the  cosine 
(cos).     Thus,  cos  AOB  =  OM :  OB ;  cos  AOD  =  OL :  OD. 

(3.)  The  perpendicular  (p)  divided  by  the  base  (h)  is  the 
TANGENT  (tan).     Thus,  tan  AOB  =  BM :  OM. 

(4.)  The  base  (h)  divided  by  the  perpendicular  (p)  is  the 
COTANGENT  (cot).     Thus,  cot  AOC  =  OK :  KG. 


5/: 


TRIGONOMETRY.  11 

(5.)  The  reciprocal  of  the  sine  is  called  the  co-secant  (csc). 
Thus,  CSC  AOB  =  OB :  BM. 

(6.)  The  reciprocal  of  the  cosine  is  called  the  Secant  (sec). 
Thus,  sec  AOD  =  OD:OL. 

Unity  minus  the  cosine  is  called  the  versed-sine  (yersin) ; 
thus  versin   AOB  =1  -  OM :  OB. 

Unity  minus  the  sine  is  called  the  coversed-sine  (coversin); 
thus  coversin  AOC  =  1  -  CK :  CO 

It  will  be  noticed: — 

(1.)  That  each  ratio  can  have  but  one  value  for  any  one 
angle.  For  if  the  perpendicular  were  dropped  from  any  other 
point  of  OB  than  B  (e.  g.,  B',  C,  D'),  the  new  triangles  of 
reference  would  be  similar  to  OBM,  and  hence  the  ratios  of 
their  sides  would  equal  the  corresponding  ratios  in  OBM. 

(2.)  Neither  the  sine  nor  the  cosine  can  exceed  1. 

(3.)  The  secant  and  cosecant  must  both  exceed  1. 

(4.)  Neither  the  versed-sine  nor  the  coversed-sine  can  ex- 
ceed 2. 

5.  The  functions  (4)  may  evidently  be  grouped  in  two  triads, 
viz.:  Sine,  Tangent,  Secant,  and  Co-sine,  Co-tangent,  Co-secant. 
Fig.  2  will  render  clear  the  reason  of  this  nomenclature.  Let 
AOB  be  any  angle,  BOQ  its  com- 
plement, BOM  and  KOL  the  tri- 
angles of  reference.  As  BOM  is  sim- 
ilar to  KOL, 

OM  :  OB  =  KL  :  OK,    or    co-sine 
AOB  =  sin  (90° -AOB), 

OM  :  BM  =  KL  :  OL,  or  co-tangent 
AOB  =  tan  (90°  -  AOB), 

OB:BM  =  OK:OL,   or  co-secant 
AOB  =  sec  (90°  -  AOB),  ,^'»-  ^' 

that  is,  the  co-sine  of  an  angle  is  the  sine  of  the  complement, 
the  co-tangent  is  the  tangent  of  the  complement,  and  the  co- 
secant is  the  secant  of  the  complement. 


12  SIGNS — LIMITING  VALUES. 

^  6,  While  each  function  has  but  one  value  for  any  one  angle, 
the  converse  is  not  true.  To  a  single  numerical  value  of  a 
function  may  correspond  four  different  angles,  each  less  than 
360°.  Any  confusion  that  might  result  from  this  fact  is 
avoided  by  the  adoption  of  the  following  convention  (Cf.  Alg. 
§§8,  46,  55,  429,  432) :~ 

Vertical  distances  are  estimated  from  the  initial  line  XOX' 
(Fig.  1) ;  those  measured  upward  are  said  to  he  positive;  hence 
those  measured  downward  must  be  taken  as  negative.  Hori- 
zontal distances  are  estimated  from  the  vertical  line  YOY'; 
those  laid  off  to  the  right  are  said  to  he  positive;  hence  those 
measured  to  the  left  are  negative.  The  hypothenuses  of  the 
triangles  of  reference  being,  in  general,  neither  vertical  nor 
horizontal,  may  be  considered  signless. 

Consequently  all  functions  of  angles  in  the  first  quadrant 
are  positive.  For  angles  in  the  second  quadrant  the  base  of 
the  triangles  of  reference  are  negative,  and  the  perpendiculars 
positive ;  hence  the  sine  and  cosecant  are  positive^  and  the  other 
functions  negative.  In  the  third  quadrant  both  base  and  per- 
pendicular are  negative;  hence  the  tangent  and  cotangent  are 
positive,  and  the  other  functions  negative.  Finally,  for  angles 
in  the  fourth  quadrant  the  cosine  and  secant  are  positive,  and 
the  other  functions  negative. 

7.  Limiting  Values  of  Functions.  Tlie  Sine.  At  0° 
p  in  the  triangle  of  reference  is  evidently  0,  and  hence  sin  0°  = 
0 :  /i  =  0.  It  will  be  convenient,  though  not  necessary,  to  assume 
that  the  length  of  h  remains  constant  as  the  angle  increases 
from  0°  to  360°;  let  the  changing  angle  be  denoted  by  x.  As 
X  increases  from  0°  onward,  p  increases,  and  hence  sin  x  in- 
creases, rapidly  at  first,  but  slowly  as  x  approaches  90°. 
When  X  is  nearly  90°,  p  and  h  are  nearly  equal,  and,  finally, 
when  X  =  90°,  p  and  h  coincide,  and  p  :  /*=  1,  .-.  sin  90°  =  1. 
As  X  increases,^  (and  hence  p :  h)  decreases,  slowly  at  first,  but 


TRIGONOMETRY.  13 

more  rapidly  as  x  approaches  180°.  When  x  is  very  near 
180°,/)  (and  hence ^:  h)  is  very  small;  finally  when  re  =  180°, 
p:  h  =  0.  A  peculiar  circumstance  here  presents  itself.  As 
180°  =  180°  -  0°,  and  also  180°  +  0°  (Alg.  Ch.  VIII)  the  angle 
of  180°  may  be  considered  as  ending  in  either  the  second 
quadrant  or  the  third.  In  the  first  case  sin  180°  is  +,  and  in 
the  second  case-.  Obliged  to  recognize  both  contingencies 
we  say  that  sin  180°  =±0.  Here,  then,  sin  x  changes  its 
sign,  becoming  negative.  As  x  increases,^,  and  therefore  p :  h, 
increases  numerically  until,  when  a;  =  270°,  p  and  h  coin- 
cide, and  p:  h=  -1.  An  x  increases,  p  (and  hence  p :  Ji)  de- 
creases numerically,  until,  when  x  =  360°,  sin  x  =  +  0  according 
as  X  is  taken  as  360° -0°  or  360° +  0°.  If  the  angle  gen- 
eratrix continue  to  revolve,  the  same  values  will  evidently 
recur  in  the  same  order. 

The  Tangent  When  a;  =  0°,^  =  0,  and  .-. p:b  =  0,  i.  e.,  tan  0° 
=  0.  As  a;  increases  p  increases  and  b  decreases  at  the  same 
time;  hence  tan  x  increases  very  rapidly.  When  x  is  nearly 
90°,  tan  X  is  very  great,  and,  when  x  =  90°,  tan  x  =p ;  0  =  oo  . 
As  X  may  at  this  stage  be  considered  as  ending  in  the  first 
(90° -0°),  or  in  the  second  (90°  +  0°),  quadrant,  and  as  the 
tangent  is  positive  in  the  first  and  negative  in  the  second,  we 
say  that  tan  90°  =  ±od  .  Here  the  tangent  changes  sign  by 
passing  through  qo  .  As  a:  increases  toward  180°  tan  x  rapidly 
diminishes  (numerically),  since  p  diminishes  and  b  increases, 
until,  when  a;  =  180°,  tan  x^T  0.  Tan  x,  changing  its  sign 
by  passing  through  0,  increases  rapidly  as  x  increases  toward 
270°.  When  a;  =  270°,  tan  .t=  tec  ,  and  as  x  increases,  tan  x, 
changing  its  sign  by  passing  through  co ,  decreases,  numeri- 
cally, with  great  rapidity,  until,  when  a;  =  360°  tan  a;=+  0. 

A  consideration  of  the  changes  in  all  the  functions  as  x 
changes  from  0°  to  360°  leads  to  the  results  given  in  the  fol- 
lowing 


14  "FUNDAMENTAL   RELATIONS." 

►:  Table  of  Limiting  Values  of  the  Functions. 


Function. 

0°  to  90° 

90°  to  180° 

180°  to  270° 

270°  to   360° 

Sine 

Oto+1 
+ 1  to±0 

0  to±oo 
4-  ooto±0 

1  t0±00 

C0t04l 

4-1  to±0 
±0  to-1 
±ooto+0 
±0  toqrx 
±x  to-1 
+  1  to±oo 

±0  to-1 
-1    to  +  0 
+  0  to±c» 
T  CO  to±0 

- 1    toqroo 
±  00  to  -  1 

-1  to  +  0 

Cosine 

+  0  to  4-1 

Tangent  

Cotangent 

Secant 

±00  to  +  0 
±0  to+co 

Cosecant 

-1  to^co 

8.  Negative  Angles.  By  comparing  the  functions  of 
negative  angles  (2)  with  the  functions  of  numerically  equal 
positive  angles,  it  will  be  seen  that,  A  being  any  angle, 
sin  (  -  J.)  =  -  sin  J. ;  esc  (  -  J.)  =  -  esc  A;  cos  (  -  J.)  =  cos  J. ; 
sec  (  —  A)=  sec  A;  tan  (  -  J.)  =  -  tan  A;  cot  (  -  ^)  =  -  cot  A. 
Hence  changing  the  sign  of  an  angle  changes  the  sign  of  the 
SINE,  COSECANT,  TANGENT,  and  COTANGENT,  but  does  not  af- 
fect the  sign  of  the  cosine  and  secant. 

9.  Fundamental  Relations.*  If  A  represent  any  angle, 
(1.)  Sin'  A  +  cos'^  ^  =  1.  (4.)  Tan  ^  cot  JL  =  1. 

sin  A  (5.)  Sin   A  esc  A  =  1. 

cos  A  (6.)  Cos  A  sec  A  =  1. 

i     cos  A  (7.)  Sec'  A  =  l+  tan'  A. 

~"      "■        •  (8.)  Csc^ 


(2.)  Tan  A 
(3.)   Cot 


sin^         •  (8.)  Csc'^  =  l  +  cot'X 

Of  these,  (2),  (3),  (4),  (5),  (6),  follow  immediately  from  the 

P  ^ 

definitions.     To  prove  (1):  Sin  A  =  -/  cos  A  =  -  hence  sin'  A  + 

«  h  h 

p'     6'    p'+¥     h'  ,      .    M    , 

cos'  A  =  —  -{-  —  = =  --  =  1.    (7)  and  (8)  may  be  similarly 

/t'     /i'         /r        h 

established. 

10.  Anti-functions.     If  ab=.m,  we  write   b  =  a-hn.     By 

analogy  we   usually  say  that   if  tan  A=m,  J.  =  tan~^m,.  al- 


*The  student  must  memorize  these,  and  be  thoroughly  familiar  with  them. 


TRIGONOMETRY. 


15 


though  the  word  tangent  cannot  be  considered  a  factor. 
Though  this  notation  can  be  justified,  it  is  inconvenient  and 
may  produce  confusion,  as  powers  of  functions  are  indicated 
by  placing  the  exponent  above  and  a  little  to  the  right  of  the 
abbreviation  for  the  function  ;  thus  (tan  A)^  is  usually  written 
tan'  A  ;  so  that  tan""^  A  might  mean  "t^"  as  well  as  "  the  angle 
whose  tangent  is  A."  As  it  is  useless  now  to  attempt  to 
change  the  notation,  the  student  must  be  on  his  guard  to  avoid 
confusion.  While  sin~^  A,  cot~^  A,  etc.,  are  sometimes  read  "  anti- 
sine  of  tI,"  "  inverse  sine  of  J.,"  etc.,  it  is  better  to  read,  in 
full,  "  the  angle  whose  sine  is  A,"  etc. 

11.  The  trigonometric  functions  may  be  represented  by  lines, 
if  it  be  assumed  that  all  angles  are  formed  about  the  centre  of 
a  circle  of  unit  radius,  and  that  in  every  triangle  of  reference 
one  side  is  equal  to  the  radius.  In  this  case,  however,  the 
angles  are  treated  by  means  of  the  measuring  arcs,  and  it  is 
better  to  consider  the  functions  as  functions  immediately  of  the 
arcs.  Arcs  considered  directly  are  generally  supposed  to  be- 
gin at  the  right  hand  extremity  of  a  horizontal  diameter  (and 
we  shall  assume  that  they  do  so  begin);  this  point  is  called  the 
primary  origin,  or  simply,  the  origin.  The  complements  of 
arcs  are  in  this  case  estimated  from  the  upper  extremity  of  a 
vertical  diameter ;  this  point  is  called  the  secondary  origin. 

The  sine  is  the  perpendicular 
dropped  from  the  termination  of 
the  arc  on  the  diameter  through 
the  origin.  Thus  sin  AP  (Fig. 
3)  is  PM.  The  cosine  of 
an  arc  is  the  sine  of  the 
complement;  thus  cos  AP  = 
PL;  cos  AD  =  KD.  The  tan- 
gent is  that  part  of  the  geo- 
metric tangent  a(  the  origin 
which  is  limited  by  the  diame- 
^^'  '•  ter  through  the  termination  of 


r. 


1(5  TRIGONOIk^ETRIC  LINES. 

the  arc;  thus  tan  AP  =  tan  ABF  =  AT.  The  cotangent  of  an 
arc  is  the  tangent  of  the  complement ;  thus  cot  AD  =  BE. 
The  secant  is  the  line  connecting  the  centre  of  the  circle  with 
the  extremity  of  the  tangent ;  thus  sec  ABD  =  OS.  The  co- 
secant of  an  arc  is  the  secant  of  the  complement ;  thus  esc  AP 
=  0V.  The  versed-sine  is  the  radius  minus  the  cosine,  and 
the  co-versed-sine  is  the  radius  minus  the  sine ;  thus  versin  AP 
=  MA ;  coversin  AD  =  KB. 

Lines  are  always  measured /ro?^  A  A'  and  BB';  those  meas- 
ured upward  and  those  measured  to  the  right  of  BB'  are  pos- 
itive ;  all  other  horizontal  or  vertical  lines  are  negative.  The 
secant  and  cosecant  follow  in  ^ign  their  reciprocals,  the  cosine 
and  sine,  respectively.* 

If  the  radius  be  different  from  unity  the  lines  indicated 
above  must  be  divided  by  the  radius,  in  order  to  get  the  various 
functions.  As  these  definitions  agree  in  substance  with  those 
given  in  (4)  we  may  use  angles  or  their  measuring  arcs  in- 
differently, and  likewise  the  trigonometric  functions  or  their 
line  representatives. 

12.  Functions  of  SO''  and  of  ^5°. 

Let  AOB  (Fig.  4)  =  30°.     Then  (Ven. 

36  p.  51)  OB  =  2  AB, .-.  sin  AOB  = 

AB :  OB  =  ^.     As  sin'  30°  +  cos'  30^ 

=  1,  cos''  30°  =  1  .-.  cos  30°  =  ^  Vs. 

Fig.  4^  T,an_30°=i--- 

J  V^3  =  Jl/'S;  cot  30°  =  1/3;  sec  30°  = 

f  1/3 ;  CSC  30°  =  2. 

Again  let  AOC  (Fig.  5)  =  45°.  As 
OC  is  the  diagonal  of  a  square,  AC : 
OC=sin  45°  =  il/2=cos  45°.     As OA  ^ig.  5.  ^ 

=  AC,  tan  45°  ^  cot  45°  =  1 .     Finally,  sec  45°  =  esc  45°  =i/2. 
13.  Trigonometric  Tables. — To  enable  us  to  deal  with 
angles  by  means  of  their  functions  tables  are  computed,  giving 
the  values  of  these  functions  for  all  angles  from  0°  to  90°. 

*The  versed-sine  and  coversed-sine  are  measured  from  the  foot  of  the  sine  and 
cosine  respectively,  and  therefore  are  never  negative. 


TRIGONOMETRY.  17 

The  angles  are  usually  taken  at  intervals  of  V,  though  the  in- 
terval may  be  as  large  as  10'  in  tables  for  ordinary  use, 
and  is  frequently  as  small  as  1"  in  tables  used  for  calculations 
requiring  great  accuracy.  The  functions  of  an  angle  lying 
between  any  two  angles  of  the  table  are  found  by  interpola- 
tion (Alg.  491  and  Appendix  III). 

14.  The  Tables  of  Natural  Functions  give  the  actual  val- 
ues of  the  functions ;  the  Tables  of  Logarithmic  Functions 
give  the  common  logarithms  of  these  values.  As  all  sines  and 
cosines,  tangents  of  angles  between  0°  and  4^°,  and  cotangents 
of  angles  between  45°  and  90°,  are  "  proper "  fractions,  the 
characteristics  of  their  logarithms  are  negative.  To  avoid 
these  negative  characteristics,  10  is  added  to  the  characteristic 
of  the  logarithm  of  each  of  these  functions,  and,  in  many  tables, 
for  uniformity,  10  is  added  to  every  characteristic. 

15,  Elementary  Method  of  Computing  a  Table  of 
Natural  Functions. — We  will  consider  the  line  representa- 
tives of  functions,  i.  e.,  (11)  the  functions  of  the  arc  (radius 
being  unity).  Though  an  arc  exceeds  its  sine,  the  inequality 
diminishes  as  the  arc  decreases,  the  ratio  (sin  x)  :  x  having  for 
its  limit  1.  This  inequality  is  so  small  for  an  arc  of  V  that 
there  is  no  error,  within  the  limits  of  ordinary  tables,  in 
assuming  arc  l'  =  sin  1'.  In  a  circle  of  radius  1,  the  semi- 
circumference  =  10800' =  TT  =  3.14159265;  hence  arc  1'  = 
3.14159265^10800  =  .000290888-sinr.    cos  V  =  ^/l~mi'V 

=  V  (1+sin  l')(l-sin  T)  =-/l.000290888  x  .999709112 
=  .999999958 
It  will  be  shown  (37)  that  if  A  and  B  be  any  two  angles, 

(I)  sin   (^  +  J5)  -  2  sin  ^  cos  J5  -  sin  {A  -  B). 

(II)  cos  {A  +  B)  =  2  cos  ^  cos  i>  -  cos  {A  -  B), 

Let  B=  V,  and  let  A  become  successively,  V,  2',  3',  etc. 
From  (I),  sin  (1'  + 1')  =  sin  2'  =  2  sin  1'  cos  V 
Sin    (2'  +  V)  =  sin  3'  =  2  sin  2'  cos  1'  -  sin  V 
sin  4'=  2  sin  3'  cos  1'  -  sin  2', 
and  so  on. 


18  TRIGONOMETRIC   TABLES. 

.     From  (II)  COS  2'  =  2  cos'^  1'  -  / 

cos  3'--  2  cos  2'  cos  V  -  cos  1' 
cos  4'=  2  cos  3'  cos  V  -cos  2' 
and  so  on. 

As  2  cos  1'=  1.999999154  =  2 -.000000846  is  a  constant 
factor  in  the  first  term  of  the  second  member,  these  equations 
may  be  written 

sin  2'-  2  sin  1'  -  .000000846  sin  V  =  .000581776 

sin  3'  =  2  sin  2' - .000000846  sin  2' -sin  1'  =  . 000872665 
sin  4'  =  2  sin  3'  -  .000000846  sin  3'  -  sin  2'  =  .001163553 
and  similarly  for  the  cosines.     After  computing  the  functions 
of  angles  up  to  30°  by  these  formulae,  we  let  A  (I)  remain 
equal  to  30°,  while  B  changes  from  V  successively  at  minute 
intervals  up  to  15°.     Then,  as  sin  30°  =  ^,  2  sin  30°- 1,  and 

we  have 

sin  (30°  + 1')  =  cos  1'  -  sin  29°  59' 

sin  (30°  +  2')  =  cos  2'  -  sin  29°  58' 
sin  (30°  +  3')  =  cos  3'  -  sin  29°  57' 
and  so  on  up  to  45°,  the  only  operation  required  in  the  second 
members  being  mere  subtraction. 

Instead  of  (II)  we  now  use,  (37),  the  formula 

cos  (A  +  B)^  -  2  sin  ^  sin  ^  +  cos  (A  -  B). 
Letting  A  remain  equal  to  30°,  we  have 

cos  (30°  +  ,1')  =  -  sin  1'  +  cos  29°  59' 
cos  (30°  +  2')  =  -  sin  2'  +  cos  29°  58' 
and  so  on,  up  to  45°  inclusive. 

We  need  not  compute  beyond  45°,  as  the  sine  and  cosine  of 
any  angle  between  45°  and  90°  are  respectively  the  cosine  and 
sine  of  the  complement,  i.  e.,  of  an  angle  between  45°  and  0° 
(of  which  the  cosine  and  sine  have  been  computed). 

The  tangents,  cotangents,  secants,  and  cosecants  may  be 
obtained  from  the  sines  and  cosines  by  means  of  the  funda- 
mental relations  (9).  Functions  of  angles  greater  than  90° 
may  be  obtained  from  the  table  by  means  of  the  relations 
suggested  in  (5,  6). 


TRIGONO]NIETRY.  19 


16.  The  Table  of  Logarithmic  Functions  may  be  obtained 
by  simply  tabulating  the  common  logarithms  of  the  natural 
functions  and  adding  10  to  each  characteristic.  Other  and 
better  methods  will  be  indicated  later.  This  table  is  usually 
employed  in  preference  to  the  table  of  natural  functions,  as 
logarithms  are  utilized  whenever  they  serve  to  facilitate  com- 
putation, as  in  multiplications,  divisions,  involution,  and  evolu- 
tion. 

17.  Tables  are  usually  so  arranged  that  for  angles  from  0^ 
to  45°  the  degrees  and  the  names  of  the  functions  are  to  be 
taken  from  the  top  of  the  page,  and  the  minutes  from  the  left- 
hand  column,  while  for  angles  between  45°  and  90°  the 
degrees  and  the  names  of  the  functions  are  to  be  taken  from 
the  bottom  of  the  page,  and  the  minutes  from  the  right-hand 
column. 

18.  Interpolation  is  effected  on  the  hypothesis  that  as  t]ie 
arc  changes  uniformly,  the  function  changes  uniformly.  The 
error  involved  in  this  assumption  being  greatest  for  cosines  of 
very  small  angles  and  sines  of  angles  near  90° ^  we  avoid  com- 
puting small  angles  by  means  of  the  cosine,  and  angles  near 
90°  by  means  of  the  sine,  unless  we  have  tables  specially  cal- 
culated for  the  purpose.  It  is  best  in  general  to  compute 
angles  by  means  of  the  tangent  or  cotangent. 

19.  To  guard  against  the  errors  likely  to  creep  into  the 
extended  computations  of  (15)  we  check  the  results  of  those 
computations  by  comparing  them  with  the  values  of  functions 
of  various  angles,  determined  independently  by  formulae  that 
might  in  this  connection  be  termed  verification  formulse.. 
Numerous  examples  of  these  will  be  given  later. 


I 


CHAPTER  II. 

SOLUTION"    OF    TRIANGLES. 

The  parts  of  a  triangle,  trigonometrically  considered,  are 
its  sides  and  angles.  Plane  Trigonometry  treats  in  detail  the 
problem : — "  Given  three  parts  of  a  triangle,  to  compute  the 
other  three,"  or,  more  generally,  "From  data  sufficient  to 
determine  a  plane  triangle,  to  compute  the  unknown  parts." 
Of  given  parts  one  at  least  must  fix  the  length  of  a  line;  angles 
only  can  determine  nothing  more  than  the  shape  of  a  tri- 
angle : — the  ratios  of  the  sides  to  each  other. 

In  any  triangle,  ABC,  the  angles  will  be  denoted  by 
A,  B,  C,  and  the  sides  opposite  by  a,  6,  c,  respectively. 

20.  Right  Triangles. — The  problems  that   arise  in   the 
solution  of  right  triangles  may  be  grouped  thus : — 
I.  Given  the  hypothenuse  and  another  side. 
II.  Given  the  two  sides  about  the  right  angle. 

III.  Given  the  hypothenuse  and  an  acute  angle. 

IV.  Given  an  acute  angle  and  a  side,  opposite  or  adjacent. 
The  definitions  (4)  may,  (Fig.  6) 

be  stated  thus: — 

(1)  sin  C  =  c^  a 

(2)  cos  C=h^a 

(3)  tanC  =  c-^  h 

(4)  cot  Q=h^G 
These  suffice  for  the  solution  of  all 

right  triangles.  ^'°"  ^' 

I.  1°.  Given  a,  h,  (2)  gives  cos  C,  whence  0  is  obtained  at 
once  from  the  tables,  c'^  =  d^ -  h^ ;  hence  c  =  V {a->rb)  (a-  h) 
and  the  solution  is  complete,  c  might  also  be  computed  from 
(1)  after  C  has  been  found;  but,  (i)  any  error  in  the  computa- 
tion of  0  would  vitiate  this  computation,  and  (2)  if  C  be  near 
90°  its  sine  cannot  be  accurately  determined  from  the  usual 
(20) 


TRIGONOMETRY.  21 

tables.  Required  parts  should  always  be  determined  directly 
(if  possible)  from  given  parts.  Unused  relations  may  then 
serve  as  "  checks  "  on  the  work. 

2°.  Given  a,  c.     Solution  similar  to  that  just  indicated. 

II.  Given  h,  c.  From  (3)  and  the  Tables,  C  and  hence  B 
are  determined  at  once.  To  determine  a  we  have  the  two  for- 
mulae a=l/6^  +  c^  and  a  =  c-rsin  C;  the  former  is  direct,  but 
not  adapted  to  logarithmic  computation ;  the  latter,  the  simpler, 
has  the  disadvantage  already  indicated. 

III.  Given  b,  B.  C  =  90°  -  B.  (2)  gives  a,  and  (3)  deter- 
mines c.     When  c,  C  are  given  the  solution  is  similar. 

21.  Oblique  TrianorleS. — As  every  oblique  triangle  may 
be  divided  into  two  right  triangles,  all  oblique  triangles  may 
be  solved  by  (20). 


Fig.  7.  Fig.  8. 

I.  Given  two  sides  and  the  included  angle,  e.  g.,  c,  h,  A. 
Draw  from  the  extremity  of  one  known  side,  (6),  CD  J_  c,  the 
other  known  side.  Solve  ACD  (20,  HI),  thus  determining 
k  and  p;  if  ^  >•  c,  CD  falls  without  the  triangle  (Fig.  7). 
BD  =  G—  k  or  k-  c,  according  as  CD  falls  within  or  without. 
Knowing  p  and  m,  solve  the  triangle  BDC,  thus  obtaining  a 
and  BCD.  ACB  =  ACD  +  BCD  or  ACD -BCD  according 
as  CD  falls  within  or  without  ACB. 

II.  Given  two  angles  and  a  side,  e.  g..  A,  B,  h.  All  the 
angles  are  known,  and  the  side  6.  Draw,  from  the  extremity 
of  this  side,  CD  J_  c.  Solve  ACD  (I).  Then,  knowing  p 
and  the  angle  DCB,  solve  DCB.     If  CD  falls  mthin  the  tri- 


22 


SOLUTION    OF   TRIANGLES. 


angle,  i.  e.,  if  ACD  <  C  (as  given),  AB  =  AD  +  DB.     If  CD 
falls  without,  i.  e.,  if  ACD  >  C,  AB  =  AD- DB. 


Fig.  9.  Fi?.  10. 

III.  Given  two  sides  and  the  angle  opposite  one  of  them, 
e.  g.,  h,  a  A.  From  the  extremity  of  a  known  side,  h,  draw 
CD  _L  c,  the  unknown  side.  Solve  ACD,  as  before.  Know- 
ing |)  and  a,  solve  DCB  (20, 1).  As  in  this  case  two  triangles 
may  fulfill  the  given  conditions,  it  is  called  the  "ambiguous 
case"  Thus  if  a  <  6,  a  may  take  either  one  of  the  positions 
CB  or  CB\  cutting  c  at  equal  distances  from  the  foot  of  the 
p3rpendicular,  and  both  ACB  and  ACB'  fulfill  the  require- 
msnts.  If  b  =  a,  ACB'  shrinks  to  a  line,  and  there  is  but  one 
solution.  Again  if  a  >■  6,  a  can  fall  on  only  one  side  of  AC 
and  there  can  be  but  one  solution.  If  a=p,  ACB'  and  ACB 
coincide.     Finally,  if  a  <  ]:>  there  is  no  solution. 

IV.  Given  the  sildi  a,  b,  c.  Drop  the  parpendicular  from 
any  vertex,  as  C.     Then  (Fig.  7), 

p'=.b'-k'  =  a^-^m' 
.'.  6'—  x^  =  Ic^  -  m\  whence 
(6  -  a)(6  +  a)  =  (^  -  ?n)(^ -f  »i),  i.  e., 
b  +  a:  k  +  m: :  k-m:  b-a     (1); 
^  (6  +  «)(6-a) 


or  k 


Determining  k-m  from  (2),  we  have 

A;  =  J(A; -H  m) -f  J(^  -  7rt),  and 

m  =  ^(k  -f-  m)  -  \{k  -  m). 


TRIGONOMETRY.  23 

Knowing  k  and  m,  we  solve  ACD  and  DCB,  and  thus  de- 
termine A,  B,  C. 

22.     The  solutions,  I,  III,  (21)  may  be  abbreviated  by 
noting  that  "  the  sides  of  a  plane  triangle  are  proportional  to 
the  sines  of  the  opposite  angles"     Thus  (Figs.  9,  10), 
p  =  bsin.A  =  a  sin  B,  whence  a:  6  =  sin  A  :  sin  B. 

This  proposition  may  be  used  to  "  check  "  solutions  of  (2.1). 

The  constant  ratio = —  = is    the   modulus   of 

sni  A     sin  B      sin  C 

the  triangle,   and   is    numerically  equal   to    the  diameter  of 

the  circumscribing  circle. 

NUMERICAL    EXAMPLES. 

1.  Right  Triangles.— 1.  Let  a  =  256,  5  =  172.  From 
(2)  cos  C  =  sin  B  =  172  -^  256  =  .67188.  The  table  of  Natural 
Functions  gives  C  =  47°  47'  14",  and  hence  B  =  42°  12'  46". 
€  =  V  {ct  +  b)(a-h)=  l/428x84  =  l/35952  =  189.6.  (1)  now 
serves  as  a  "  check,"  to  test  the  correctness  of  the  computation. 
Thus  sin  47°  47'  14"  =  .74065,  .-.  c  =  256  x  .74065  =  189.6  as 
before. 

2.  Let  6  =  .0795;  c  =  .0386  [Logarithmic  solution].  From 
(3),  tan  C  =  c-7-6,  whence  log  tan  C  =  log  c-log  6  +  10  = 
2.58659-  2.90037  +  10=  9.68622,  whence  C=  25°  53' 26". 
From  (1),  a  =  c  +  sin  C,  .*.  log  a  =  log  c-log  sin  C  +  10  = 
2.58659  -  9.64025  + 10  =  2.94634  /.  a  =  0.08838.  The  rela- 
tion a"^=  b'^  +  c"^  may  be  used  as  a  "  check." 

3.  Let  a  =27.125,  C=  37°  42' 18".     From   (1), 
€=a  sin  C 

log  a=- 1.43337  Again,  6  =  a  sin  B;  log  a  =  1.43337 
log  sin  C=  9.78647  log  sin  B  =  9.89827 

log  c=  1.21984  log  ^>  =  1.33164 

c=  16.589  6=21.46 

IL  Oblique  Triangle8.-l°.Given  A=56°41',  B=74°  28' 


24  NUMERICAL    EXAMPLES. 

^and  6=1326.  C=180°- (A  +  B)  =  48°  51'.  InACD(Fig.7) 
AD=  b  cos  A  and  CD=  6  sin  A. 

log  1326=3.12254  log  1326=  3.12254 

log  cos  56°  41'  =  9.73978         log  sin  56°  41'  =  9.92202 

log  AD  =  2.86232  log  CD  =  3.04456 

.-.AD  =728.31 
In   CDB,  CB  =  CD  -^  sin  B,  and  DB  =  CD.cotB. 

log  CD  =  3.04456  log  CD  -  3.04456 

log  sin  74°  28'  =  9.98384         log  cot  74°  28'  =  9.44397 

log  CB  =  3.02840  log  DB  =  2.48853 

a=  1067.59  DB  =  307.99 
c=  AD +  DB=  1036.3 

We  check  our  results  by  the  law  of  sines  ;  thus : 
sin  A  -^  a  should  equal  sin  B  -^  6. 

log  sin  B-  9.98384  log  sin  A=  9.87679 

log  ^>=  3.12254  log  a=  3.01549 

6.86130  6.86130 

Thus  the  logarithms  of  the  two  ratios  agree  exactly. 

2.  Given  a=  1200,  b=  758,  A=  72°  14'.     As  a>  5  there 
is  but  one  solution. 

In  ACD,  CD=  5  sin  A,  and  AD=  b  cos  A. 
log  sin  72°  14'=  9.97878         log  cos  72°  14'=  9.48450 
log  758=  2.879(37  =2.87967 

log  CD=  2.85845  log  AD  =T3G4r7 

CD=  721.8  AD=  231.3 

In  CDB,  BD=  |/(a  +  CD)(a  -  CD)  =  v^  1921.8  x  478.2 
log  1921.8=  3.28371      log  sin  B=  log  CD  -  log  a  +  10 
log  478.2=2.67961  log CD=  2.85845 

2  I  5.96332  log  a=  3.07918 

log  BD=  2.98166  log  sin  B=  9.77927 

BD=  958.6  B=  36°  58' 49" 

ACB=  180°  -  (A  +  B)  =  70°  47'  11". 

3.  Given  a  =  12.76;  6  =  26.58,  c=31.6.      Then,  (Fig.  7), 


TRIGONOMETRY.  25 

,            39.34x13.82     ,^^,      i     oa  4  no 

k-m= — =  17.21 .-.  ^'=24.4;  w^7.2. 

ol.D 

log  COS  B=log  7.2  -  log  12.76  + 10 

=.85733  -  1.10585  +  10  =  9.75148.-.  B=55°38'  57". 
log  cos  A=log  24.4  -  log  26.58  4- 10 

=1.38739  -  1.42455  +  10=  9.96284  .-.  A=  23°  22' 
C=  180°  -  (A  +  B)  =  100°  59'  3".     As  a  check  we  may  use 
(22)  ;  thus  :—    log  b  =  1.42455  log  c=  1.49969 

log  sin  B  =9.91677  log  sin  C=  9.99197 

log  modulus=  1.50778  or,  1.50772 

results  agreeing  to  the  fourth  decimal  place. 
4.  Given  6=.0079,  c=  .0124,  A=  37°  22'  17". 
In  ACD  (Fig.  7), 

log  sin  A  =  9.78318  log  CD  =  3.68081 

log  .0079  =  3.89763  log  DB  =  3.78675 

log  CD  =  3.68081  log  tan  B  =  9.89406 

.-.      B  =  38°4'48" 
log  cos  A  =  9.90021  log  CD  =  3.68081 

log  .0079  =  3.89763  log  sin  B  =  9.79012 

log  AD  =  3179784  log  a  =  3.89069 

AD=   .00628  a=   .00777 

.-.  DB=   .00612 
We-may  again  check  by  use  of  (22),  thus: — 

loga=  3.89069  log  b=  3.89763 

log  sin  A  =9.78318  log  sin  B  =9.79012 

log  modulus=  2.10751  log  modulus  =2. 107 51 

results  agreeing  exactly. 


CHAPTEK  III. 

FUNCTIONS  OF  THE  SUM  AND  OF  THE  DIFFERENCE 
OF  TWO  ANGLES.  MISCELLANEOUS  FORMUL.F. 


23.  Sine  and  Cosine  of  the  Sum. — To  show  that 

(1)  sin  (^x  +  2/)  =  sin  x  cosy  +  cos x  sin  y 

(2)  cos     ^y')^cosx  cos  y-sinx  sin  y. 

Let  OA,  OB,  and  OC  be  any  three  lines  whatever,  forming 
the  angles  AOB  and  BOC  (Figs.  11,  12).  Let  AOB-  x  and 
BOC=-  y.  From  any  point  of  OC,  as  F,  drop  FE  _|_  OA  and 
FP  j^  OB,  and   from   P   drop   PD  j_  OA  and  PG  _[_  EF. 


Fig.  11.  Fig.  12. 

Then  PD  :0P  =  sin  x  =PG:  FP  (similar  triangles),  and 
OD :  0P=  cos  a;=  FG :  FP ;  sin  y^  FP :  OF,  cos  y=0¥:  OF, 
sin  (x  +  y)  =  FE :  OF,  and  cos  (a;  +  y)  =  OE :  OF. 
From  the  figures,  EF  :  EG  +  GF  =  PD  +  GF,  i.  e., 


EF^PD  PO     FG  FP    . 
OF~  PO  •  OF  "^  FP  *  OF '  ^*  ^'' 

8m(x  +  y)  =  sin  X  cos  y  +  cos  X  sin  y 
Again,  (Fig.  11),  OE  =  OD-GP,  whence 
OE^OD  OPGP  FP    . 
OF~OP*OF     FPOF'*'^' 
cos(a;  +  y)  =  cos x  cosy- sin x  sin y 
(26) 


(I). 


(II). 


TRIGONOMETRY.  27 

The  same  formulae  may  be  deduced  from  figure  12,  if  it  be 
noted  that  in  this  case  OE  is  negative. 
24.  Sine  and  Cosine  of  the  Difference. — To  show  that 

(1)  sin  (x-y)^  sin  x  cos  y  -  cos  x  sin  y 

(2)  cos  (x-y')  =  cos  x  cos  y  +  sin  a  sin  y. 


Fig.  13.  Fig.  U. 

Let  AOB  (Fig.  13)  be  x,  and  BOC  (laid  off  backward,  i.  e., 
in  a  negative  direction),  be  -  y,  so  that  AOC  is  x-y.  From 
any  point,  D,  of  OC.  draw  DH  _l  OA,  and  DE  _l  OB. 
Draw  EG  i  OA,  and  DF  x  EG.  Then  EG :  OE  =  FD  :  ED 
=  di\x,  OG:OE  =  EF:ED  =  cosa;,  DE:0D  =  sin2/,  OE:OD 
=  cos  y,  DH :  OD  =  sin  (x  -  y),  and  OH :  OD  =  cos  (x  -  y). 
From  the  figure  we  see  that  DH  =  EG  -  EF,  whence 

DHEG  OEEF  ED   . 

OD"OE'OD     EDOD'"'^' 

sin  (a;  -y)=  sin  x  cosy-  cos  x  sin  y  (HI)* 

Again,  OH  =  OG  +  FD,  whence 

ohog  oe   fd  ed  . 
od~oeod"^ed'od''-^" 

cos  (a;  -y)  =  cos  x  cos  y  +  smx  sin  y      •  (I^« 

In  Fig.  14,  both  x  and  x-y  are  obtuse. 

mil  f Of^  -I-  7/1 

25.  Tanirents. — Since  tan(a;  +  w)=  — \ ^,  we  obtain, 

cos  {x  +  y) 

by  substitution  from  (31),  tan  (x  +  y')  = — — - 

cos  X  cos  y  -s\n  X  sm  y, 

Tlie    student    should  notice   that  no  supposition  has  been  made  concerning  the 
direction  of  OB  and  OC,  and  that,  therefore,  the  formulae  are  true  for  all  angles. 


28 


MISCELLANEOUS   FORMULAE. 


or,  dividing  both  numerator  and  denominator- by 'CO&^c-cosy, 

(remembering  that =  tan  x,  etc.), 

cos  a; 


tan  {x  +  y) 


_     tan  X  +  tan  y 
1  -  tan  X  tan  % 


(V). 


Bya  similarprocess we^obtain  tan  (x  -  y)  =~.^ ~  (VT)- 

-'  ^  \       :/y     1-f  tana:tan  y  ^      ^ 

cos  \  X  "4"  \l^ 

34.    Cotangents. — Since    cot(a;4-v)  =— r— ^ — =^,  weob- 

^         sm(a;  +  2/) 

*  •     V       u  i-i.  i.-        /otN       x/         N     COS  a;  COS  w  -  sin  a;  sin  y 

tam,- by  substitution-  («>1)>  cot(a;  +  v)  =  -; ; — ^• 

sin  a;  cos  y  +  cos  x  sin  y* 

or,    dividing    both    terms    of    the     fraction     by  sin  x  sin  y, 

cot  a;  cot  2^  -  1 


cot  (a3H-i/)  =  - 

cot  2/  +  cot  a; 

Similarly  we  obtain  cot  (a^  -  y)  = 


cot  a;  cot  3/  -  1 


(VII). 
(VIII). 


cot  2/  -  cot  a; 

35.  Exercise.— Find  (31  -  34)  the  functions  of  (90±a:), 
(180°±a;),  (270°±a;),  (360°±a;).  The  results  are  given  in  Table 
II.  Observe  that  if  the  number  of  quadrants  involved  be 
odd  the  function  changes  name,  and  not  otherwise.  Thus^ 
tan(270°  +  a;)=  -cota;;  but  tan(180°  +  aj)  =tanaj.  The  sign 
is  determined  by  the  quadrant  in  which  the  angle  would  lie, 
if  a;  were  less  than  90°. 

Table  II. 


Function. 

90^  +x 

90* -X 

180°+a; 

ISC-x 

270°+x 

270°— X 

360° +x 

360°-x 

Sink. 

cos  X 

cos  X 

—sin  X 

sin  X 

—cos  X 

—cos  X 
—sin  X 

sin  X 

—sin  X 

Cosine. 

-sin  X 

sin  X 

—cos  X 

—cos  X 

sin  X 

cos  X 

cos  X 

Tangent. 

—cot  X 

cot  X 

tan  X 

—tan  X 

—cot  X 
—tan  X 

cot  X 

tanx 

—tan  X 

Cotangent. 

—tan  X 

tan  X 

cot  X 

—cot  X 

tan  X 

cot  X 

—cot  X 

Secant. 

—CSC    X 

CSC  X 

—sec  X 

—sec  X 

CSC    X 

—esc  X 

sec  X 

sec  X 

Cosecant. 

sec  X 

sec  X 

—CSC    X 

CSC    « 

-sec  X  —sec  x 

CSC   X 

--CSC   X 

36.  Functions  of  Double  Angles  and  Half  Angles. 

If  in  (31-4)  we  make  y=  x,  we  obtain, 


TRIGONOMETRY.  29 

sin  t^+"^)  =sin  (x  +  x)=  sin  2  :r=  2  sin  x  cos  x  (A) ' 

cos  (x  +  y)  --cos  (.i;  +  ^)  =  cos  2x=  cos'^  a;  -  siu^  x 

=2cos''a;-l=l-2sin2a;  (B) 

/         N     ^      o  2  tan  ^  ^i-,N 

tan  (x  +  x)=  tan  2  a;=- ■ —  {yj 

1  -  tan''  X 

cot^ic-1  ^■n>. 

cot  (x  +  x)^  cot  z  a;=  — r K^J 

^  ^  2  cot  a; 

Again,  from  (B),  we  have  1  -  cos2a:=  2  sin^a;.     Let  a; 

=  -^,  whence  2  x=a.     Then  2  sin'  -=  1  -  cos  a,  whence  sin  ^ 
2  2  2 

=  ^"1(1  ~  ^o^  ^0  (^')-     From  (B)  we  have,  also,  2  cos'''  x= 

1  +  cos  2  X,  or,  if  a;=  -^,  2  cos'^  ^=1  4-  cos  a,  whence  cos—  = 
2  2  -^ 

l/j(l  +COS   a)  (B').    From  (A')  and  (B')  we  derive,  at  once, 

a         /l  -  cos  a   .^,.        ,      ,  ^         /l  +  cos  a       ^-p.,. 

tan  -  =^/- (C);  and  cot  ^  =-y/- (D  ). 

2      \  1  -f-cos  a  ^      \  1  -  cos  a 

These  are  formulae  of  great  importance. 

28.  Adding  (I)  and  (III),  member  to  member,  we  obtain 
sin  {x-\-y)-\-  sin  (a?  -  ?/)  =  2  sin  x  cos  y  (1).  Let  x-\-y  =  m  and 
a;  -  2/ =  ^,  whence  ic  =  |^(?/i  + /:)  ;  2/  =  i(^'i'- ^)-  Then  (1)  be- 
comes sin  m  +  sin  ^•  =  2  sin  \{in  +  A;)  cos  ^(m  -  ^)  (2); 

Similarly, 

sin  m  -  sin  ^  =       2  cos  \{iin  +  ^)  sin  |(?/i  -  U)  («J); 

cos  m  4-  cos  ^  =       2  cos  |^(m  +  ^)  cos  \{m  -  k')  (4); 

cos  m  -  cos  ^  =    -  2  sin  ^(r^i  +  U)  sin  J(m  -  ^)  (5). 

29.  From  (2) -(4),  (28)  we  deduce,  by  direct  division, 

sin  m  +  sin  k _2  sin  ^(m  +  k)  cos  \{m -  k~) _ tan ^(m  +  ^)    ,^ ^ 

sin  m- sin  ^     2cos^(m  +  ^)  sin  J(m-A;)     tan^(m-^)        ' 

and  similarly, 

cos?)i-cosA;         ^       1/     .  7^x       i>        7N  r9^. 

— i  =  -  tan  *(wr+  A^)  tan  -J^(m  -  A;)  K.'^jf 

cos  ?7i  +  cos  A;  "^  ^ 

sinm  +  sinA;      ^      ,,        zn  /on  sinm-sin  k  i\/a\ 

=  tan  Urn  +  k)  (3):  j-  =-cot|(m—  A;)  (4). 

cosm  +  cosA;  ^         oosm-cosA; 

30.  The  formulae  of  (28,  29)  enable  us  to  adapt  formulae 
involving  sums  or  differences  of  functions  to  logarithmic  calcu- 
lation. 


CHAPTER   lY. 

ADDITIONAL  FORMULA  FOR  PLANE  TRIANGLES. 

40,  A7iy  side  of  a  plane  triangle  equals  the  sum  of  the  prod- 
ucts of  the  remaining  sides  into  the  cosines  of  the  respective 
angles  included  between  them  and  the  first  side. 

Thus  (in  Fig.  7),  if  CD  i  AB,  AD  =6  cos  A,  and  BD  = 
a  cos  B,  whence  c  =  AD  +  DB  =  b  cos  A  +  a  cos  B.     If  (as  in 
Fig.  8),  CD  fall  without  the  triangle,  c  =  AD  -  DB.    But  DB  = 
a  cos  CBD  =  a  cos  (180°  -  B)  =  -  a  cos  B 
and  we  have  as  before,  c=  6  cos  A  +  a  cos  B  (1) 

Similarly  6  =  c  cos  A  +  a  cos  C  (2) 

a  =  ^  cos  C  +  c  cos  B  (3) 

41,  The  square  of  any  side  of  a  triangle  equals  the  sum  of 
the  squares  of  the  other  two  sides,  minus  twice  the  product  of  these 
sides  into  the  cosine  of  their  included  angle. 

From  (40),  (2),  by  transposing  and  squaring,  we  obtain 
d^  cos"^  C  =  5^  -  2  6c  cos  A  +  c^  cos'^A 

From     (22)  a'^sin^C= c^sin^A 

Adding,  a'  =  b'' -2  be  cos  A  +  c'  (1) 

Similarly,   *  h'  =  a'  +  G'-2ac  cos  B  (2) 

c'=b''  +  d'-2abcosO  (3) 

42,  From  (1)  -  (3),  (41)  we  deduce 

cosA  = 

cosC  =  ^^±^^'(3) 
2  be 

These  formulae,  not  being  adapted  for  logarithmic  computa- 
tion, are  usually  transformed  as  follows  : — 
(30) 


2  be        ^^ 


TRIGONOMETRY.  31 

Subtracting  each  member  of  (1)  from  luiity,  we  have 

,      ,      b^  +  c'-a:'     2cb-b'-c'  +  ci:'     ci'-(b-cy 
l-eosA=.l ^^— = ^^ ==-^^- ;    or, 

since  l-cosA^2sm-^4'  2sin^^i  =  (.^zA±£^^ 

Let  2  8=a  +  b  +  c;  then  s—b^^(a  +  c- b),  and  8-c  =  ^(a  +  b-c'), 
so  that 


sm     ~^-      ^X-e) 


.   A        Ks-bX' 


■^    (2);  VR^       rtrr    ^Hft-  r 


similarly,        «n-  =  y -^^ (2);  WB)      r^ 

^^"2  =  V-^;i—     ^^>j 

Adding  both  members  of  the  equations  (A)  to  unity  we 
obtain,  by  similar  reductions,  formulse  (C). 


cos 


cos 


|B  =  ^!(i2_*)  (2);  1(C)         ^-^«^     '^f'^ 


/.■^  (.s  -  c) 
ab 


cosiC  =  y^^l^^  (3). 


From   (B)   and  (C)   we   obtain,  by   direct    division,   (D). 
tan 


tan 


tan^C=^/^l-;i^^^  (3). 


(s-a)(s- 


Of  these  three  sets,  (B),  (C),  (D),  the  first  should  be  used 
only  when  the  angles  are  small;  the  second  should  never  be  used 


32 


SPECIAL    FORMULA. 


with  small  angles.  The  formulse  (D) 
are  the  best  for  general  itse,  being 
accurate  for  all  angles.  These  form- 
ulae furnish  the  means  of  solving  Case 
IV  (21). 

34.  Radius  of  Inscribed  Circle 
(r).  Area.— Let  ABC  be  any  triangle. 
The  bisectrices  of  the  angles  meet  in 
the  centre  of  the  inscribed  circle. 
Also  AE  =  s-CB,  BD  =  s-AC,  etc. 
Hence,  in  the  right  triangle  AOE, 

tanOAE  =  tan^A--=-^  = 
s  -a 


-  a 


=  (33,  D), 


Us-b)(is-c)_    1      / 
\      sijs  —  a)         s-a\ 


(s—a)(s-  bXs-  c) 


=  tan  ^A.{s-a 


)-^/o^-«x.^- 


bXs-c) 


Again,  as  the  area  of  a  triangle  equals  one-half  the  peri- 
meter multiplied  by  the  radius  of  the  inscribed  circle,  the  area 
of  ABC 


V(s—a')(s--b)(s  —  c)       /—. ^, TTT : 
^^: <-^ ^-^ ^  =  -^ s  {s  —  a){s—b){s ~- g). 

35.  Another  expression  for  the  area  is  "one-half  the  prod- 
uct  of  any  two  sides  into  the  sine  of  the  included  angle." 

Thus  (Figs.  7,  8),  area  ABC  =  ^  AB .  CD  =  J  AB .  6  sin  A  = 
■|  be  sin  A.     This  is  equivalent  to  be  sin  ^  A  cos  J  A. 

36 .  The  sum  of  any  two  sides  of  a  triangle  is  to  their  dif- 
ference as  the  tangent  of  half  the  sum  of  the  opposite  angles  is 
to  the  tangent  of  half  their  difference. 

Let  ABC  be  the  triangle.  By  (22),  a:b  ::  sin  A:  sin  B. 
By  composition  and  division,  a  +  b:a-b:  :  sin  A  +  sin  B : 

sin  A  -  sin  B  :  :  (29),  tan  i(A  +  B)  :  tan  |(A  -  B).     Q.  E.  D. 

Case  II  (21)  may  be  solved  by  means  of  this  proposition. 


CHAPTER   V. 

SPHERICAL  TRIGONOMETRY. 

36.  Spherical  Trigonometry  treats  of  the  solution  of 
spherical  triangles,  or  the  relations  between  the  face-angles 
and  diedrals  of  the  triedrals,  whose  common  vertex  is  the 
centre  of  the  sphere,  and  which  intercept  these  triangles  on 
the  surface.  As  the  sides  of  these  triangles  are  expressed  in 
degrees,  or  in  radial  measure,  the  solution  of  the  triangle  gives, 
not  the  absolute  lengths  of  the  sides,  but  only  the  ratios  ex- 
isting between  them. 

To  determine  the  absolute  lengths  w^e  must  know  the  radius. 
This  premised,  any  three  parts  of  a  spherical  triangle  suffice 
to  determine  the  other  three.  Two  parts  are  of  the  same 
species  when  they  lie  in  the  same  quadrant,  and  of  diflferent 
species  w^hen  they  lie  in  different  quadrants.  We  consider 
only  triangles  in  which  no  part  exceeds  180°;  this  convention, 
though  not  absolutely  necessary,  simplifies  our  calculations 
without  causing  any  material  loss  of  generality. 

37.  Right  Triangles.  Napier's  Parts.  Napier's 
Rules. — ^To  express,  in  two  general  rules,  the  various  formulae 
required  in  solving  a  right  triangle,  Napier  adopted  the  follow- 
ing convention :  In  any  right  triangle  the  two  sides  about  the 
right  angle,  the  complements  of  the  two  angles  opposite  them, 
and  the  complement  of  the  hypothenuse,  are  called  the  five 
circular  parts.  If  any  three  of  these  be  considered,  one 
will  be  adjacent  to  both  the  others,  or  one  will  be  separated 
from  each  of  the  others  by  the  intervention  of  a  circular  part. 
A  part  adjacent  to  two  others  is  called  a  middle  part,  and  the 
other  two  are  called  adjacent  extremes.  A  part  separated 
from  two  others  is  called  a  middle,  and  the  other  two  are  called 

opposite  extremes. 

(33) 


34 


NAPIER  S   EULES. 


Napier's  Rules  —I.  The 

SINE   of   the  middle   part 
equals  the    product  ,of  the 

COSINES     of   the     OPPOSITE 

extremes. 

Taking  the  circular  parts 
successively  as  middles,  we 
Fig.  15.  have    the    following  equa- 

tions to  establish : — 

(1)  sin  J  =  sin  a  sin  B 

(2)  sin  c  =  sin  a  sin  C 

(3)  cos  a  =  cos  h  cos  c 

(4)  cos  B  =  cos  h  sin  C 

(5)  cos  C  =  cos  c  sin  B 

Let  ABC  be  the  right  triangle,  A  being  the  right  angle. 
Connect  the  vertices  with  the  centre  (O)  of  the  sphere.  From 
any  point  D,  of  OC,  drop  DE  ±  O  A,  and  from  E  draw  EF  x  OB. 
Then  DF  must  be  perpendicular  to  OB,  and  DFE,  the  recti- 
lineal angle  of  the  diedral,  OB,  will  equal  the  spherical  angle 
B.  DEO,  EOF,  DOF,  and  DFE,  are  respectively  triangles 
of  reference  for  h,  c,  a,  and  B. 

(1)  As  sin  i  =  DE  :0D,  sin  a  =  DF :  OD,  and  sin  B  =  DE  :DF, 
we  have,  at  once, 

DE_DF  DE 
0D~uI3  DF' 
sin  h  =  sin  a  sin  B.     Q.  E.  D. 
The  lettering  of  the  figure  being  purely  arbitrary,  so  far  as 
B  and  C  (and  therefore  b  and  c),  are  concerned,  (2)  follows  at 
once  from  (1). 

OF^OE  OF 

OD     ODOE'  ' 

we  have,  at  once,  cos  a  =  cos  h  cos  c.    Q.  E.  D. 

(4)  To  r.how  that  cos  B  =  cos  6  sin  C.     We  cannot  get  sin  C 

sin  c 
directly  fro::a  the  %ure,  so  we  replace  it  by  -: —  (2),  as  sin  c  and 


or 


(3)  As 


TRIGONOMETRY.  35 

sin  a  can  be  obtained   directly.     We   must  prove,  then,  that 

„      sin  c  . 

cos  ±>=  cos  0—. ;  1.  e.,  cos  i3=  cos  o  sin  c  esc  a. 

,     sin  a 

^  ^       ,         ,.       ,     EF     OE  EF  OD     . 
But  we  have,  directly,  j^^q^  "  OE  '  FD  '  ^'  ^•' 

cos  B  =  cos  h  sin  c  esc  a  =  cos  h  sin  C.  Q.  E.  D. 

II.  The  SINE  of  the  middle  part  equals  the  product  of  the 
TANGENTS  o/  the  ADJACENT  extremes. 

We  are  to  prove  that : — 

(1)  sin  6  =  tanc  cotC 

(2)  sin  c=  tan  h  cot  B 

(3)  cos  B  =  tan  c  cot  a 

(4)  cos  C=  tan  h  cot  a 

(5)  cos  a  =  cot  B  cot  C 

(1)  From  (I,  (1))  we  have,  sin  6  =  sin  a  sinB;  from,-(I,  (2)) 

siaa=  -. — ;^,and  from  (1,(5))  sinB=  ;  whence 

sm  C  ^  ^  ^^  cos  c 

.     ,      sin  c   cos  C     sin  c  cos  C  /-.  ^  -r^  t^ 

sm  b  =  ■-:—r- . = -: — —  =  tan  -c  cot  C  Q,  E.  D. 

sm  O  cos  c     cos  c  sm  C 

(2)  Similarly  sin  c  =  tan  h  cot  B. 

(3)  From  (1,(4))  cos  B=  cos  6  sin  C,  and  as 

,      cos  a         -    .     „     sin  c        , 

cos  0= ,  and  sm  C=  -. — ,we  have 

cos  c  sin  a 

^     cos  a  sin  c    ^  ^  >^  -r^  t^ 

cos  B  = : —  =tan  c  cot  a.     Q.  E.  D. 

cos  G  sin  a 

(4)  Similarly,  cos  C=  tan  h  cot  a. 

cos  B 

(5)  Finally,  as  (I)  cos  a=  cos  6  cos  c,  and  cos  h=—. — ^, 

sin  v^ 

,  cos  C  cos  B  cos  C        ,  T^        ^    ^   T^   T^ 

and  cos  c^-. — =^,  cosa=-^ — -,.-^ — ^=cotB  cotC.  Q.  E.  D. 
sm  B  sm  C  sin  B  ^ 

88.  Napier's  Rules  suffice  for  the  solution  of  all  problems 
involving  only  right  triangles.  Slight  difficulties  may,  however, 
present  themselves  in  actual  practice,  when  we  wish  to  decide 
whether  an  angle  (arc)  whose  functions  are  numerically  deter- 
mined, lies  in  the  first,  or  in  the  second,  quadrant. 


36  DETERMINATION   OP   SPECIES. 

38.  Determination  of  Species.— If,  in  any  investiga- 
tion, a  part  be  determined  by  means  of  the  cosine,  tangent,  or 
cotangent,  the  sign  of  the  function  will  indicate  the  qu^-drant 
in  which  the  part  is  to  be  found.  If,  however,  the  part  be 
determined  by  means  of  the  sine,  an  ambiguity  presents  itself, 
as  the  sine  is  positive  in  both  the  first  and^  second  quadrants. 
In  this  event  we  note  that : — 

I.  An  oblique  angle  and  the  side  opposite  are  always  of  the 
same  species. 

From  (37)sinC= p    as  sinC  is  necessarily  positive, 

cos  B  and  cos  b  must  agree  in  sign ;  i.  e.,  B  and  b  must  lie  in 
the  same  quadrant. 

II.  If  the  hypothenuse  exceed  90°,  the  remaining  sides,  and 
consequently  the  oblique  angles,  are  of  different  species. 

From  (37),  cos  a  =  cos  b  cos  c.  Hence,  if  a>  90°,  cos  a  is 
negative,  and  therefore  cos  b  and  cos  c  are  of  different  signs ; 
i.  e.,  b  and  c,  and  therefore,  by  (I),  B  and  C,  lie  in  different 
quadrants.  If,  however,  a  <  90°,  cos  a  is  positive,  and  hence 
cos  b  and  cos  c  agree  in  sign,  and  therefore  b  and  c  (and  like- 
wise B  and  C),  lie  in  the  same  quadrant.     Hence 

III.  If  the  hypothenuse  be  less  thayi  90°,  the  remaining  sides, 
and  therefore  the  oblique  angles,  are  of  the  same  species. 

IV.  If  a  side  {other  than  the  hypothenuse'),  and  the  angle 
opposite  be  given,  there  will  be  two  solutions,  if  the  sine  of  the 
side  be  less  than  the  sine  of  the  angle;  one  solution  if  the  sine 
of  the  side  equal  the  sine  of  the  angle,  and  no  solution  if  the 
sine  of  the  side  exceed  the  sine  of  the  angle. 

From  (87),  sin  a=-^— — -;  hence    (4)  sin  b  cannot  exceed 
^      ^  smB  .     ^ 

sin  B.     If  sin  6  =  sinB,  sin  a  =  \,  a  =  90°,  and  the  triangle  is 

birectangular.      If  sin  b  <  sin  B,  sin  a  <  1,  but  positive,  and 

a  may  lie  either  in  the  first  or  in  the  second  quadrant. 

40.  Quadrantal    Triangles,  i.   e.,  triangles  in  which  at 

least  one  side  is  a  quadrant,  may  be  solved  by  Napier's  Rules, 


TRIGONOMETRY.  37 

as   the   polar  of  every  quadrautal  triangle  must  be  a  right 
triangle. 

41  i  Example. 

In  a  right  triangle,  6=  38°  16';  c=  120°  35';  solve. 

From  (37)  cos  a=cos  h  cos  c,  whence 

log  cos  a=  log  cos  6  +  log  cos  c  -  10 

=  9.89495  +  9.73121  -  10=  9.62616. 
h  <90°  and  c>  90°;  .-.  cos  6   is  +,  and  cos  c  is  -,  so  that, 
cos  a  is  negative,  and  a>  90°;  .*.  a=  155°  0'47".     To  avoid. 
using  a  we  compute  C  by  (37,  II,  1),  which  gives 
log  cot  C=  log  sin  6  -  log  tan  c 

=  9.79192  -10.19442 +  10  =  .  5975. 
e  >  90°;  C  >  90°,  (89)  /.  C=  111°  35'  40". 
Finally,  log  cot  B=  log  sin  c  -  log  tan  b 

=  9.92563-  9.89697  =.02866 
whence  B=  43°  06'  40". 

42.  The  formulae  of  Spherical  Trigonometry  correspond 
closely  to  those  of  Plane  Trigonometry  ;  differences  arise  from 
the  fact  that  the  sides  of  plane  triangles  are  dealt  with  directly, 
while  the  sides  of  spherical  triangles  are  dealt  with  only  by 
means  of  their  functions.  Analogies  deserve  attention  so  far 
as  they  aid  in  fixing  formulae  in  mind.  Table  III  suggests 
analogies  between  the  formulae  for  solving  Plane  Right  trian- 
gles and  those  employed  in  solving  Spherical  Right  triangles. 
More  striking  analogies  will  present  themselves  in  the  case  of 
oblique  triangles. 

Table  III. 

Higfit  Plane  Triangles.  JUgJa  Spherical  Triangles. 


sin  B=  -  ;  sin  C  = 

a* 

c 

a  ' 

•    T)      sm  * 

sm  B= ; 

sm  a 

.    ^      sm  c 

smC=  -: . 

sm  a 

cosB=-;    cosC  = 

b 

a ' 

^      tane 

cosB  =  - : 

tan  a 

^      tan  5 

C08C=-T . 

tana 

tanB  =  -;  tanC  = 
c 

c 

b' 

^      T>      tan  6 
tanB=-^ — ■; 
sm  c 

^      ^      tan  c 

tanC=  -T— ,. 

sm6 

COtB  =  ^;    cotC  = 

b 

c  ' 

^  -r,      sin  c 

cotB==   -• 

tan  6 

^  ^       sin  b 

cotO=-- . 

tan  c 

38 


OBLIQUE   TRIANGLES. 


Fig.  17. 


'"  43.  Oblique  Triangles. — I.  Given  two  sides  and  the  in- 
cluded angle,  e.  g.,  6,  c,  A.  From  the  extremity  of  one  known 
side  drop  BD,  perpendicular  to 
the  other.  Solve  (37)  ABD.  If 
AD  •<  b,  the  perpendicular  falls 
within  the  triangle ;  if  AD  >•  b, 
the  perpendicular  falls  without.  In 
either  case,  p  and  m  being  known, 
BCD  may  be  solved.  Finally, 
the  angle  ABC  =  ABD  +  DBC,  or 
ABD  -  DBC,  according  as  DB 
falls  within  or  without  the  triangle. 

II,  Given  two  angles  and  the  in- 
cluded side,  e.  g.,  B,  A,  c.  Drop 
the  perpendicular,  BD,  from  one 
extremity  of  the  known  side,  to  b.  Solve  the  triangle  ABD 
as  in  I.  If  ABD  >•  B  the  perpendicular  falls  without ;  if 
ABD  <  B,  the  perpendicular  falls  within  the  triangle,  p 
and  DBC  being  known,  the  triangle  DBC  may  be  solved. 
Finally,  AC  =  k  +  m,  or  k-m,  according  as  BD  falls  within  or 
without  ABC. 

III.  Given  two  sides,  and  an  angle  opposite  one  of  them; 

e.  g.,  a,  c,  A.  Drop  BD  i.  b, 
the  unknown  side.  Solve  ABD 
(37);  in  DBC  the  sides  a  and 
BD  will  then  be  known,  and 
DBC  may  be  solved.  But  as 
two  equal  arcs  may  be  drawn 
from  a  point  to  an  arc  of  a  great 
circle,  a  may  occupy  either  of 
two  positions,  one  on  the  right, 
and  the  other  on  the  left,  of  BD. 
Hence  (compare  (21)  III),  there 

^^s-'^s.  ujay  \yQ  t^YQ  solutions.     There 


TRIGONOMETRY.  39 

will  evidently  be  two  solutions  if  a  be  intermediate  in  value 
between  jo  and  c,  and  aho  between  p  and  180°  -  c. 

I.  If  A  <  90°,  a>p,  and  < c,  but  a  >  180°  -  c,  there  can 
be  but  one  solution,  as  ABC";  for,  though  a  might  occupy  a 
second  position  BC",  as  far  to  the  right  as  BC"  is  to  the  left, 
of  BD,  C"  would  fall  beyond  A',  since  BC"  >  BA',  and  the 
arc  AC"  would  exceed  180°,  so  that  we  could  not  (36)  con- 
sider the  triangle  ABC"'. 

II.  If  A<90°,  a>j9,  and  a>  c,  but  ^«<180°-c,  a 
must  fall  to  the  right  of  the  perpendicular,  and  there  will  be 
but  one  solution. 

III.  If  A  >  90°,  the  conditions  I,  II,  must  of  course  be 
changed  by  reversing  the  signs  of  inequality. 

IV.  If  o=i?,  ABC  and  ABC  coincide  with  ABD. 

V.  If  a  <.pj  A  •<  90°,  there  will  be  no  solution,  as,  when 
A  <  90°,  the  perpendicular  is  the  shortest  arc  (of  a  great 
circle)  from  B  to  ADA'. 

VI.  If  a  >•  JO,  A  >•  90°,  there  will  be  no  solution,  for  if 
A  >  90°  the  perpendicular  is  the  longest  arc  of  a  great  circle 
that  can  be  drawn  from  B  to  ADA'. 

The  number  of  solutions  should  be  determined  as  soon  as 
BD  is  computed. 

IV  •  Given  two  angles  and  a  side  opposite  one  of  them,  e.  g., 
A,  C,  a.  The  polar  or  supplemental  triangle  may  be  solved 
a?  in  III.  The  angles  and  sides  of  ABC  may  then  be  ob- 
tained by  taking  the  supplements  respectively  of  the  sides  and 
angles  of  the  polar. 

V.  Given  the  sides :  a,  b,  c.     From  any  vertex,  as  B,  draw 


40  OBLIQUE   TRIANGLES. 

^'BJ),J.b.     Let  AD  =  w,    and  CT>=k.     Then,    (87), 
cos  c=cosm  cos^  (1),  and  cosa=  cos  A;  cos^  (2),  whence 

cos  c :  cos  a^  cos  m  :  cos  k,  and,  by  division  and  composition, 

cos  c  -  cos  a  :  cos  c  +  cos  a^  cos  m  -  cos  k :  cos  m  +  cos  k,  or 
(29,2),  tan  i(c  +  a).tan  i(c-a)  -  tan  ^(m  +  k) .  tan  |(m  -  yt)(3). 
From  (3)  either  m+k  ovm-k  may  be  determined  if  the  other 
be  known.  If  BD  fall  within  the  triangle,  in  +  k=b;  if  BD 
fall  without,  m-k=h  (numerically).  The  symmetry  of  (3) 
shows  that  either  ^(in  +  k)  or  \(m—  k)  may  be  assumed  equal 
to  ^  6  in  the  first  instance,  the  other  being 
t^j^-i  tanKc  +  ft)  tanK^-g) 

tan  ^6  ^  ^' 

Adding  this  latter  value,  (4),  to  ^b,  we  obtain  the  greater  seg- 
ment ;  subtracting  (4)  from  ^b  we  obtain  the  other  segment. 
If  either  segment  exceed  b,  the  perpendicular  must  fall  with- 
out, and  b  must  be  the  numerical  difference  of  the  segments. 

The  signs  of  the  functions  in  (3)  are  important ; 
tan  |(c  -H  a)  will  be  +  if  a  +  c  <  180°,  and  -  if  c  +  a  >  180°  ; 
tan  ^(c  -  a)  is  +  if  e'>'  a,  and  -  if  c  <<  a ;  for  c  -  a  cannot  ex- 
exceed  1 80°.     tan  J6  is  + . 

If  the  tangent  (4)  be  negative  it  is  the  tangent  of  half  the  nu- 
merical difference  of  the  segments,  and  m  <  k.  For  tan  ^(_,n  +  k) 
is  always  + .  For,  the  feet  of  the  two  perpendiculars  that  can 
be  dropped  from  B  to  AC...  are  180°  apart,  and  if  they  both 
fall  without  the  triangle  we  take  D  as  the  foot  of  the  nearer. 
Hence  k  +  m  is  always  less  than  180°. 

k  and  m  being  determined,  ABD  and  CBD  may  be  solved, 
(37),  and  thus  all  parts  of  ABC  obtained. 

From  (3)  we  derive  the  following  important  theorem  (com- 
pare 21,  IV):— 

If  from  any  vertex  of  a  spherical  triangle,  a  perpendicular 
be  dropped  to  the  opposite  side,  the  tangent  of  half  the  sum  of 
the  segments  of  this  side  is  to  the  tangent  of  half  the  sum  of  the 
other  tivo  sides  as  the  tangent  of  half  the  difference  of  the  sides 
is  to  the  tangent  of  half  the  difference  of  the  segments. 


TRIGONOMETRY.  41 

VI.  Given  the  angles  A,  B,  C.  Solve  the  polar  by  Y.  The 
supplements  of  its  angles  are  the  sides  of  the  original  triangle. 

The  solutions  III,  IT,  may  be  facilitated  by  use  of  the  the- 
orem, "  The  sines  of  the  sides  of  a  spherical  triangle  are  propor- 
tional to  the  sines  of  the  opposite  angles." 

Thus  (Figs.  19, 20),  sin  j9=  sin  a  sin  C ;  also,  in  ABD,  sin^  = 
sin  c  sin  A,  whence  sin  a  sin  C—  sin  c  sin  A  or, 
sin  a :  sin  c : :  sin  A :  sin  C. 

This  theorem  may  also  be  used  as  a  "  check  "  on  the  other 
solutions  of  this  article. 

SECTION  II.— ADDITIONAL  FORMULA. 

4:4:,  Right  Triangles. — When  B  or  C  is  small,  computa- 
tion by  means  of  the  cosine  should  (18)  be  avoided.  We  may 
use  the  following  formula : — 

,  B     1  -  cos  B     1  -  tan  c  cot  a     tan  a  -  tan  c      sin  (a  -  c), 
2      1  +  cos  B     1  +  tan  c  cot  a     tan  a  +  tan  c     sin  (a  +  c) 

whence  tan  4  B  =  -i  /  -; — 7 ^  (I). 

^  V  sni  (a  +  c)  ^  ^ 

Again,  if  a  be  near  90°,  we  may  use  the  formula 

tan  ^  =  y  -  ^^g^gt^jx  [proof  similar  to  proof  of  (I)]       (II). 

If  6  or  c  be  near  90°,  it  will  be  better  not  to  compute  by 
means  of  the  sine. 

45.  Oblique  Triangles. — In  any  spherical  triangle. 

The  cosine  of  any  side  equals  the  product  of  the  cosines  of  the 
other  two,  plus  the  product  of  the  sines  of  these  sides  into  the 
cosine  of  the  included  angle. 

To  prove  (Figs.  16, 17)  cos  b  =  cosa  cos  c  +  sin  a  sin  c  cos  B.  (1) 
By  (C7)  cos  a  =  cos  m  cos  p;    cos  c  =  cos  k  cos  p,     \  ,on 

sin  a  =  sin  ^  ^  sin  C ;  sin  c  =  sin  />  -f^  sin  A  )  ^  ^ 

Again,  (37)  cos  B  =  cos  ABD  cos  DBG  -  sin  ABD  sin  DBC. 
By  (37)  cos  ABD  =  sin  A  cos  ^ ;  sin  ABD  =  sin  ^  -^  sin  c ; 
cos  DBC  =  sin  C  cos  m ;  sin  DBC  =  sin  m  -f  sin  a 


42  ADDITIONAL    FORMULAE. 

XT  T>  J    ...    ^     sin  m sink  ^ 

xlence  cos  B  =  cos  m  cos  k  sin  A  sin  C  -  -; : — .  (Z) 

sin  a  sin  c  ^  "^ 

Substituting  the  values  (2),(3)  in  the  second  member-of  (1), 
we  obtain 

cosm  cos  k  cos^p  +  cos  m  cos  k  sin^jy  -  sin  k  sin  m,  i.  e., 
cos  m  cos  k  -  sin  w  sin  k,  which  equals  cos  b,  or, 
cos  a  cos  c  -f  sin  a  sin  c  cos  B  =  cos  b.     Q,  E.  D. 
Il>  fall  without,  (Fig.  17),  B  =  ABD  -  DBC,  and  6  =  ^  -  m, 
and  the  reduction  is  similar  to  that  just  indicated.     For  a  and  c 
we  have 

cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  J.  (4), 
cos  c  =  cos  a  cos  b  +  sin  a  sin  6  cos  C  (5). 
46.  -From  (1),  (4),  (5),  we  deduce 
cos  a  -  cos  b  cos  c 


cos  A  = 
cos  B  = 
cos  C  = 


sin  6  sin  c 
cos  6  -  cos  a  cos  c 

sin  a  sin  c 
cos  c  -  cos  a  cos  6 


sin  a  sin  b 

These  formulae,  like  their  analogues,  (33)  are  ill  adapted 
for  logarithmic  coitiputation.  We  therefore  transform  them, 
as  follows:  From  (1)  we  deduce 

cos  a  -  cos  b  cos  c     sin  b  sin  c  +  cos  6  cos  c  -  cos  a 


1  -  cos  A  =  1 

or,      sin""^  ^  A  = 


sin  b  sine  sin 6  sine 

cos  (6  -  e)  -  cos  a     2  sin  ^(a  +  b—c)  sin  ^(a  +  c—b) 
2  sin  6  sin  e  2  sin  6  sin  e  * 


1     •       n  T  .   i  1    A      sin(s— c)sin  (s-ft) 

or,   placing  2s  =  a  +  b  +  c,  sm^ -J- A  =  —  ~.~-,-,— \ 

sin  b  sm  c 


\  sin  6  sin  c 

sinularlv,     sin  J  B  ^    /«i"  (^  7  «)  sin  (^I)  (2).  1(B) 
\  sin  a  sine  | 

sinJC  =  ^AMH5fl£5E5  (3). 
V  sin  a  sin  A 


TRIGONOMETRY.  43 

4:7.  Adding  both  members  of  (1),  (2),  (3),  (A),  to  unity 
and  reducing  in  a  manner  similar  to  that  just  shown,  we  obtain 

cos-^  ^  —  '  ^       ^ 


I  \  _     /sin  s  sin  {s  —  a] 
\      sin  b  sin  c 

T  -D        /sin  «  si 
V      smi 


sin  (s  —  b) 


b  sine 


1  r^        /sin«sin(s  -  c) 

cos  i  C=  ^  — —-4 ^^ 

\      sni  6  sm  c 

The  tabulated  cosines  are  inaccurate  for  small  angles,  and 
the  sines  are  unsatisfactory  for  angles  near  90°.  We  therefore 
obtain,  by  direct  division,  from  (B)  and  (C),  the  tangents,  (D), 
which  may  be  satisfactorily  employed  for  all  angles. 


tan|A  =  ..M"(^-"^^'"^\-^>  (1); 
\      sm«sm(«— a) 

tanJB=J!l^5IS5E5  (2);  1(D) 

taniC=  j!i^l!HllIlfEJ)  (3). 
V      sin.'?sin(«~c) 

If  A'B'C  be  the  polar  of  ABC, 

A'  =  180°  -  a,  B'  =  180°  -  6,  C  =  180°  -  c.  ) 
a'  =  180°  -  A,  b'  =  180°  -  B,  c'  =  180°  -  C.  j 
In  this  polar  then  we  have  (45) 
cos  a'  =  cos  b'  cos  c'  -  sin  b'  sin  c'  cos  A',  etc.,  or,  by  substitution, 

-  cos  A  =  cos  B  cos  C  -  sin  B  sin  C  cos  a  (1);  | 

-  cos  B  =  cos  A  cos  C  -  sin  A  sin  C  cos  b  (2);  > 

-  cos  C  =  cos' A  cosB  -  sin  A  sin  B  cos  c  (3).  ) 

From  (1;,  (2),  (3),  by  methods  similar  to  those  employed 
in  (45-6)  we  obtain  the  formulae  (A'),  and,  adapting  these  to 
logarithmic  computation,  the  sets  (B^  and  (€')>  ^^^^  which  by 
direct  division  we  obtain  (D')- 

In  all  these  formula  s  =  ^(A  +  B  +  C).  (B').  (C),  (DO,  may 
also  be  obtained  from  (B),  (C),  (B),  by  passing  directly  to  the 


44 


ADDITIONAL   FORMULAE. 


v; 


polar  triangle,  as  in  the  deduction  of  (A')  from  (A).     The 
general  remarks  on  (A),  (B),  (C),  (D),  are  applicable  also  to 

(AO,  (BO,  (CO,  (DO- 


cos  a  = 


cos  b  = 


cos  c  = 


cosB  cos  C- cos  A, 

sin  B  sin  C 
cos  A  cos  C  -  cos  B^ 

sin  A  sin  C 
cos  A  cos  B  -  cos  C 

sin  A  sin  B 

cosScos(S- A)^ 


sin-|-a  =  -*/ — _,  .  V 

\        sniBsinC 

•    1  7.         /  -  c^s  S  cos  (S  -  B) 
\        sin  A  sui  C 


sin^c 


cos 


cos 


-V 


-cosScos(8-C) 
sin  A  sin  B 

cos(S-B)cos(S-C). 
sin  B  sin  C 


^         \        cos  S  cos  OS -A) 

i.^  1.        /-  cos  (8  -  A)  cos  (8  -  C) 

cot  i-  6  =  -•  / !i-- ^-- — -^^- '' 

^        V        cosScos(S-B) 

\        cos  8  cos  (8  -  C) 


.(A') 


(B') 


/cos(S-A)co.(S-CX     1 
V  sin  A  sin  C  j  v^  / 

lc=     /cos  (8 -A)  cos  (8 -B) 
\  sin  A  sin  B 


(D') 


48.  The  analogies  obtaining  between  the  formulae  of  Spheri- 
cal Trigonometry  and  those  of  Plane  Trigonometry  are  no- 
where more  interesting  than  at  this  part  of  our  work.  We 
deduced  in  (34)  an  expression  for  the  radius  of  the  inscribed 


TRIGONOMETRY. 


45 


circle  (r)  in  terms  of  the  sides  of  a 
plane  triangle.  We  have  an  analo- 
gous expression  in  the  spherical  tri- 
angle. Let  O,  (Fig.  21),  be  the  pole 
of  the  inscribed  circle  of  ABC. 
AOF,  FOB,  etc.,  are  right  triangles; 
OAF  =  JA,  OBD  =  |B,  etc.,  and 
AE  =  AF,  BF  =  BD,  CD  =  CE. 
Hence  (36) 

^    ,      tan  OF        tan  r 
tan  ^  A 


Fiff.  21. 


tan  h  A 


sin  AF 
1 


sin(«— a) 

i 
sin  {s—a)\ 

whence  tan  r  =  -♦ 


V 


sin  (s  -6)  sin  (s-c) 


sin  8  sin  (s  -  a) 


or 


/sin  (s  -  b)  sin  (s  -  c)  sin  {s  -  a)         tan  r 


sin  (s  -  a)  sin  (.s*  -  b)  sin  (« 


sin  (s  -  a) 
-\  in  which  r  is 
inscribed 


Fig.  22. 


cotR^ 


sin  5 
the   arcual   radius  of  the 
circle,  and  s  =  ^  (a  +  6  +  c). 

Again,  let  R  =  O'A  represent  (Fig. 
22)  the  arcual  radius  of  the  cir- 
cumscribed circle  of  ABC.  In  any 
right  triangle,  as  AO'E,  we  have, 
cos  O' AE  =  cot  R  tan  AE.  By  con- 
sideration of  the  isosceles 
AO'C,  etc.,  we  find  that 
S-Cand  AE=-|C;hence 
cos  (S  —  C) 


triangles 
0'AE  = 


tan  -^  C 


=  (47,D') 


cos  (S 


V 


-cos(S-A)cos(S-B)  _ 
cos  8  cos  (S  -  C) 

-  cos  (S  -  A)  cos  (S  -  B)  cos  (S  -  C) 


49.  Napier's  Analogies.- 

JVapier',9  Analogies,  viz. : — 


cosS 
-From  (46-7)  we  readily  deduce 


46  napier's  and  delambre's  analogies. 

tan-J(A+B)_cos^(a- 6)  tan  ^{a+b)     eos^(A-B). 

cotiC      ~cosK«+^)  tan^c     ""cosiCA  +  B)  ^  '^' 

tan  i( A-B)  ^  sin  |(a  -  6)  tan  |(a-  ^)  ^ sin  |(A  -  B) 

cot  J  (J  sini(a+6)  ^  ^'        tanjc         sini(A  +  B)^  ^' 

To  establish  (1):— 

cos  J(a  -  b)=  cos  (|-a  - 1^6)  =  cos  |-  a  cos  |^  6  +  sin  ^  a  sin  ^  6; 
substituting  for  cos^a,  etc.,  their  values  (46-7),  we  obtain 


.  ,       ,  ^      cos  (S  -  C)  -  cos  S      /cos  (8  -  B)  cos  (S  -  A) 

cosi(a-6)= ^ — ;— ^^ -x  — -  .    -p   .     .^ ^ 

sinC  V  sinBsmA 

cos  (S  -  C)  -  cos S        1      ,j-,^  ,  X 

= sinC •'''''*''  ^*^)  ^''^• 

«-..    -1    1           w      ,.    cos(S-C)  +  cosS        ^  ,-^ 

Similarly,  cos  ^(a + b}  = ^ — ^~F^~ ^^*  2  ^  W* 

By  division, 
cos  -|(a-  ^  j     cos  (S  -  C)  -cos  S     sin  (S-^C)  sin^C 
cos  i(a  +  6) "~  cos  (S  -  C)  +  cos  S  ~  cos  (S  -  -|^  C)  cos  ^  C 
As  S  =  i( A  +  B  +  C),  S  -  ^  C  =  J( A  +  B),  ancTwe  have 
cos^(ct-6)  _  tanyA  +  B)     q  j.  -p 
cos  -J(a  +  b)  cot  1^  C 

The  other  equations  may  be  similarly  established. 
50 .  In  a  similar  manner  we  may  also  establish  the  truth  of 
Delambre's  Analogies,  otherwise  known  as  Gaiiss's  Equations, 
viz. : — 

^  ^        cos^C      "      cos^c 

sinKA-B)_sinK^-5). 
^  ^         cos^C  sin^c 

cosKA  +  B)_cos|(a  +  6), 
^  ^         sin^C  cos^c 

cosKA-B)_sinK«  +  6) 
^  -^         sin^^C  sin^c 

These  formulae  afford  solutions  of  cases  I,  II  (43);  the  species 
of  the  parts  should  be  carefully  determined  from  the  signs  of 
the  functions. 

Napier's  Analogies  may  be  readily  deduced,  by  division, 
from  Delambre's. 


i    UNIVERSITY 

CHAPTEIi  yi, 


ALGEBRAIC  TRIGONOMETEY. 

51.  The  Trigonometric  functions  have  thus  far  been  treated 
in  a  somewhat  restricted  way,  from  a  point  of  view  ahnost 
purely  geometric.  We  now  propose  to  show  that  the  relation 
between  the  arc  (or  angle)  and  the  function  may  be  expressed 
algebraically,  and  thus  that  Trigonometry  may  be  treated  as  a 
branch  of  pure  Algebra.  We  will  also  extend  our  discussions 
to  other  transcendental  functions  of  the  same  general  form  as 
those  with  which  we  have  been  dealing,  and  in  properties  and 
relations  closely  analogous  to  them. 

52.  To  develop  sinx  and  cosx  into  series  arranged  accord- 
ing to  the  powers  of  x,  i.  e.,  to  express  algebraically  the  relation 
between  x  and  sin  x  and  cos  x  respectively. 

Whatever  the  relation  between  x  and  sin  x,  let  it  be  expressed 
by  the  identity 

sina;  =  M  +  Na;  +  Pa;'  +  Qic'  +  Ra^*  +  .  .  .  (1) 

in  which  M,  N,  P,  Q,  etc.,  are  independent  of  x,  and  are  to  be 
determined.  After  we  have  determined  them  we  must  further 
show  that  the  series  is  convergent. 

As  sin  0  =  0,  the  second  member  of  (1)  nmst  vanish  when 
x  =  0,  and  hence  can  contain  no  term  independent  of  x ;  .*.  M  =  0. 
Again,  as  sin(-£c)=  -sin a*,  the  second  member  of  (1)  must 
change  its  sign  (without  changing  its  numerical  value),  when 
-  X  is  substituted  for  x.  Hence  it  can  contain  no  even  power 
of  X.  Further,  the  coefficient  of  the  first  term  must  be  1, 
since  for  an  infinitely  small  arc  the  ratio  (sin  .t):  .r=  1.  We 
must  therefore  modify  the  form  of  (1)  by  assuming  a  develop- 
ment of  the  form, 

smx  =  x  +  qx'  +  Sx'  +  \Jx'  + (2). 

(47) 


48  TRIGONOMETRIC    DEVELOPMENTS. 

In  the  development  of  cos  x  no  term  of  an  odd  degree  can 
occur,  for  the  development  must  remain  unchanged  when  for  x 
we  substitute  -x.  Again,  the  development  must  contain  a 
term  independent  of  x  and  equal  to  unity,  since,  when  a;  =  0 
we  must  have  cos  £c  =  1  (7).    We  assume  then, 

co^x=  1 +  RV+  TV+  W  + (3). 

R',  T',  V,  etc.,  being  independent  of  x,  and  to  be  determined. 
Since  (2)  is  an  identity  we  may  substitute  2x  for  x,  thus : — 

sin2a;  =  2sina;  cos£c  =  2x  +  8Qa;=*  +  32S:c'  +  128Ux'  + (4). 

or,  sin x  qo?,x=  x+  A Qx^  + 16 Sa:^  +  64 JJx'  + (5). 

Multiplying  (2)  by  (3),  member  by  member,  we  obtain, 
sin  xQ.o^x  =  x+((^  +  W)x^  +  (S  +  T'  +  QR'jic' 

+  (U  +  V  +  QT'  +  SR')a:^  + . . . .         (6). 
Equating  the  second  members  of  (5)  and  (6),  we  obtain, 
a;  +  (Q  +  Wy  +  (S  +  T'  +  QR')^'  +  (U  +  V  +  QT'  +  m!)x'  + 

=  a;  +  4Qx^  +  16Sa;'  +  64LV  + 

Equating  coefficients,  (Alg.  462),  we  obtain 

Q  +  R'^4Q  (a);     8  +  T'  +  QR'  =  16S  (6);  )  ,^. 
U  +  V'  +  QT'  +  SR'  =  64U(c);etc.  P    ^ 

Again,  sin^a;  +  cos^a;  =  1;  substituting  for  sin  x,  cos  x,  their 
values,  (2),  (3),  we  obtain 

1  +  (1  +  2'K')x'  +  (R'  2  +  2Q  +  2T)x' 
+  (Q"^  +  2S  +  2T'R'  +  2\')x^ 

+  (T'=^  +  2QS  +  2U  +  2V'R'  +  2WK+ =  1.       (8). 

As  this  is  an  identity  w^e  have, 

1  +  2R'  =  0  (a');  R'  '^  +  2Q  +  2T'  =  0  (h'y,  | 

Q-^  +  2S  +  2T'R'  +  2V'  =  0  (cO;  etc.  J  ^^^ 

From  (A)  and  (B)  we  obtain  Q,  R',  S,  T',  etc.,  thus:— 

1  +  2R'  =  0,  .-.  R'=-i- 

From  (a),  R'  =  3Q,  .-.  Q  =  |r'  =  -  g-  =  -  ^j 
From(6'),l-i-  +  2T'  =  0,  .-.  T'  =  -i-4=  \ 
From  (6),  8+1  +  1  =  168,    .-.      S  =  jl=     ^ 


TRIGONOMETRY.  49 

etc.,     etc. 
Substituting  the  values  of  these  coefficients  in  (2)  and  (3) 
we  obtain 

„3         „5         „7         ^9 

(A). 


«•     x' 

x'     x' 

"  7!  ^  9! 

^2        ^4 

cos^  =  l--+-- 

x'     x' 
-6!'' 8!"- 

(B). 

The  series  (A)  and  (B)  are  convergent.     For,  the  ratios  of 

xi^       x^       x^ 
the  terms  in  successive  pairs  in  (A)  are  — -^    -— ,    — -,  and  so 

^.o      4.0      D.7 

on,  and  in   (B),  — ,  — ,    ^    ,  and    on    on.      Whatever  the 
^    ^    3.4'  6.7  9.10 

value  of  X  we  may  continue  these  fractions  until  the  denomina- 
tor is,  say,  greater  than  3x^  Then  the  sum  of  the  remaining 
terms  in  each  series  will  be  less  than  the  sum  of  a  geometric 
progression  of  which  the  ratio  is  ^ ;  hence  (Alg.  Ch.  IX)  each 
series  is  convergent.  The  convergence  may  also  be  shown  for 
all  ordinary  values  of  x  by  (Alg.  500).  On  account  of  the 
rapid  convergence,  these  series  give  a  shorter  method  of  com- 
puting a  table  of  natural  sines  and  cosines  than  that  indicated 
in  Chapter  I.* 

53.  From  (A)  and  (B)  we  may  deduce,  by  the  relations  of 
(9) 

(C). 

(E). 

•       (F). 


^Elegant  methods  of  computing  Tables  of  Natural  Functions,  and  also  methods 
of  computing  directly  Tables  of  Logarithmic  Functions,  which  would  be  out  of  place, 
however,  in  a  work'^of  this  character,  may  be  found  in  Caliet's  "Tables  de  Loga- 
rithiues"  Paris,  1879. 


tancc 

—  X  ■ 

o 

^3.5 

3^5.7 

3^5.7.9 

+  •• 

cot  a; 

1 

X 

X 

"3 

x' 
5.9 

2x' 
5.7.9 

x' 

7.25.27 

secic 

=  1 

x' 
^2 

6x' 
^2^3 

'      277x« 
5  ^  2^3^.7 

+ 

CSC  a; 

=  x~ 

■^  4-- 

X 

W2 

7x' 

Six' 
2\^\b.l  ^ 

127a;^ 

2^3^5^7  ^ 

50 


THE    EXPONENTIAL   SERIES. 


54,  The  relations  expressed  in  (A) ....  (F)  might  evidently 
be  taken  as  definitions,  replacing  those  given  in  (4).  The 
trigonometric  functions  might  then  be  treated  precisely  as  any 
other  algebraic  functions  involving  infkiite  converging  series. 
We  proceed  to  the  purely  algebraic  treatment  of  these  func- 
tions, and  to  some  important  extensions  of  Trigonometry, 
viewed  as  a  department  of  pure  algebra. 

55.  The  Exponential  Series. — Let  us  first  expand  a' 
into  a  series  arranged  according  to  the  ascending  powers  of  x. 
Assume 

a-=P  +  Qa;  +  Rx^  +  Sa;^-f-Tx*  +  Va;^  + 

This  equation,  being  identical,  must  be  true  for  every  value 
of  X.     Making  ic=0  we  have  a°=P;  i.  e.,  P=l.     Hence 

a^=l+Qa;  +  Ric2-fS.'«'  +  Ta:*  +  Va;^  + 

Squaring  both  members  of  (1)  we  obtain 


+  2QR 


x^+   R^ 

+  2QS 

+   T 

(1). 

(2). 


Again,  since  (1)  is  an  identity,  we  may  substitute-  2x  for  a;; 
so  doing  we  obtain 

a^=  1  +  2Qir  +  4Ra;^  +  8Sx^  +  IGToj*  +  32Vx^  + . .  .        (3). 

Equating  the  second  members  of  (2)  and  (3),  we  obtain  the 
identity 


l  +  2Q£c  +  2R 


a;^  +  2S 
4-2QR 


x^  +  . 


+  R^ 

+  2QS 
+  T 

1  +  2Qa;  +  AUx'  +  8Sx^  +  16Ta;V  .-.-(4> 
Equating  coefiicients  in  (4),  we  obtain, 

2R  +  Q2=  4R,  whence  R  =  -|-' 
2S  +  2QR  =     8S,  whence  S=  ^'  --=  %; 

24     4!* 
etc.,  etc. 
Substituting  the  values  thus  found  in  (1),  we  obtain, 


R'  +  2QS  +  T=  16T,  whence  T= 


TRIGONOMETRY.  51 

«'=l  +  Qx  +  |!x^+|V  +  ^jX'  +  .....  (5), 

the  required  series,  in  which  Q,  independent  of  a,  depends 
only  on  the  value  of  a.  Since  x  may  have  any  value  what- 
ever, let  ic=l :  Q.     Then 

a^  =  a^=<^  =  l  +  l+i  +  ^j  +  ^  + (6). 

The  second  member  of  (6)  is  e,  the  Natural  Base,  (x\lg. 
511);  hence  a}'-^==  e,  or  e^=a;  i.  e.,  Q=lna,  so  that 

a^  =1  +  (In  a)x  +  (In  a)^|^+ (In  a)^|^+ (In  a)*^J  + .  .  .  .     (7). 

56.  In  (7),  (55),  let  a=e ;  we  then  have 

„^  X  OC         X         X  r-i^ 

e«=l  +  «+-  +  -+^+.-+ ^^> 

As  (1)  is  an  identity,  xV  -  1   may  be  substituted  for  x. 
Making  this  substitution,  denoting  V  -  1  by  i,  we  obtain 

.^  ,        XXtXXlilL  /■c\^ 

^=^+"'-2-- 4! +3!+ 5! -6!- ^^^^ 

_         x"^     X*'     x^     x^ 
"2""^4!~6!'^8!"**' 
a?     x""     x' 

or,  by  (52), 

e^'  =  cos  cc  +  i  sin  x  (3). 

Similarly  we  may  show  that 

e~^*  =  cos  x  —  i  sin  x  (4). 

Combining  (3)  and  (4)  we  obtain, 

e--  +  e— ■  =  2  cos  aj ;  .-.  cos  x  =  ^(e''  +  e'^O  (5). 

e"'  _  e--  =  2i  sin  .r ;  /.  sin  x  =  lie"'  -  e""')  (6). 

These  relations  may  also  be  expressed  thus : — 

^xi  _^  ^-xi  =  cos  ic  +  COS  (  -  x').  (7). 

^xl  _  ^-xi  ^  ^  gjjj  ^  _  ^'  gii^  (^  _  x).  (8). 

(7)  and  (8)  are  called  "Euler's  Equations," 

57.   The  trigonometric  functions  are,  then,  merely  expo- 


52  FUNDAMENTAL   RELATIONS.      LIMITING  VALUES. 

nential  fanctions  of  the  arc,  and   may,  therefore,  be  thus 
defined : — 

8inx=l-.(e''-0  (l);tan.=  .^j,7;"l^  (3); 

,     cosa!  =  i(e-  +  e-'')  (2); -cotx  =  lg^^^  (4); 

2  2i 

^'''^=-pTP'''  ^^^'    ^^^^=^^rr^  (6). 

All  the  formulae  established  with  the  old  definitions  may 
now  be  established  analytically  with  these,  so  that  Trigonome- 
try may  be  considered  merely  a  department  of  Algebra. 

58.  Fundamental  Relations. — Multiplying  (3)  and  (4) 
(56),  we  obtain 

1  =  cos'^ic  +  sin^a;. 

This  may  also  be  obtained  by  direct  substitution.  Again, 
squaring  the  fractions  defined  as  tan  x  and  sec  a;,  and  subtract- 
ing the  first  square  from  the  second,  we  obtain 

sec^a;  -  tan^a;  =  1 .  Similarly, 

csc'^a;  —  cot'^a;  =  1. 
The  other  relations  are  mere  restatements  of  definitions. 
59  .  Limiting  Values. — By  substituting,  successively, 
|-,  r,|r,  2',  etc.,  for  x,  in  (1)  .  .  .  (6),  (57),  we  find  the  same 
set  of  limiting  values  as  given  in  the  Table  on  page  14. 

60.  Again,  by  substituting  for  x  in  (5),  (6),  (56),  ^tt  -  x,  re- 
membering that,  by  equations  (3),  (4),  when  a;=i-,  e^  =  i, 
and  e~^^  =  -i,  we  shall  obtain, 

sin a;=  cos  (^-—x);  and  similarly, 

tan  aj=  cot  (J-  —x),  and  esc  x=  sec  (|--  -  x),  as  in  (5). 

61.  The  addition  and  subtraction  formulae  of  Chapter  III 
may  be  established  by  direct  substitution  of  the  values  of  sin  x^ 
cos  a;,  etc.,  as  given  in  the  definitions  of  (57).  Or,  we  may 
proceed  thus : — 

e(.x+y)i  _  COS  (ix  +  y)  +  i  sin  (x  +  xj)  (a),  by  (3)  (56). 

But 

(Xi+yi  ^  ^i^yi  ^  ^^.^g  x-\-i^\\\  x)  (cos  y  +  i  siu  y)  (Jb). 

Equating  the  values  (a)  and  (6)  we  obtain 


TRIGONOMETRY.  63 

(cos  X  +  {  sin  X)  (cos  y-\-i  siny)  =  cos  (a;  +  y)  4-  i  sin  (ic  +  y), 
or,  expanding  in  the  first  member, 

cos  cc  cos  y  -  sin  a;  sin  y  +  %  (sin  a;  cos  y  +  cos  ic  sin  y)= 
cos  (a;  +  2/)  +  i  sin  {x  +  ?/). 

As  the  real  parts  of  the  two  members  of  the  identity  must 
be  equal, 

cos  (a;  +  y)  =  cos  a;  cos  y  -  sin  a;  sin  2/  (1). 

As  the  coefficients  of  i  in  the  two  members  must  be  equal, 

sin  (a;  +  y)  =  sin  x  cos  y  +  cos  a;  sin  y  (2). 

Since  y  may  be  negative  in  (1)  and  (2),  the  formulae  for 
cos  {x  -  y)  and  sin  (a;  -  y)  are  established.  From  these  we 
derive  at  once  the  formulse  for  tan  (x±y'),  cot  (x±y\  and  the 
remaining  formulae  of  Chapter  III. 

62.  Demoivre's  Theorem.— In  (3),  (4),  of  (56),  sub- 
stitute nx  for  X ;  there  results 

f,nxi  ^  (.Qg  nx  +  i  sin  nx ;  e~"^  =  cos  nx-i  sin  nx. 
As  e"^  =  (/'')"',  and  e~"^  =  (e~^^y,  we  obtain,  on  substituting 
for  e^^  and  e~^  their  values,  (56), 

(cos  x±i  sin  a;)"  =  cos  nx±i  sin  nXy  Demoivre's  Theorem. 

63  .  It  is  clear  that  in  the  functions  we  have  been  consider- 
ing X  is  not  necessarily  real.  On  the  other  hand  it  is  plain 
that  we  may  define  and  discuss  a  series  of  exponential  func- 
tions, analogous  to  those  already  treated,  but  having  the 
exponents  real.  From  these  we  may  develop  a  theory  as 
complete  and  interesting  as  that  of  the  ordinary  trigonometric 
functions.  These  latter  have  those  relations  to  the  circular  arc 
of  length  X,  which  we  have  already  discussed,  (11),  whence 
they  are  usually  called  circular  functions.  The  functions, 
the  treatment  of  which  we  are  about  to  undertake,  have  anal- 
ogous relations  to  the  arc  of  the  equiangular  hyperbola,  whence 
they  are  called  hyperbolic  functions.  Taking  again  the  natu- 
ral base,  e,  we  define  as  follows : — 

the  hyperbolic  sine  of  x,  (sinhx)  =  o(^  -  6~^)  O-y^ 


54  DEFINITIONS THE    HYPERBOLIC    FUNCTIONS. 

%  1 

the  hyperbolic   cosine  of  x,  (coshx)^ -^(e^ +  e~^)     (2) 

the  hyperbolic  tangent  of  x,  (tanh  x)  =  ^^  (3) 

e^  -T  €■ 

the  hyperbolic  cotangent  of  x,  (coth  a?)  = (4) 

2 

the    hyperbolic   secant  of  x,  (sech  x')  = — ■  (5) 

2 

the  hyperbolic  cosecant  of  x,(cschx)= — -  (6). 

From  these  definitions  we  derive  at  once  series  and  relations 
analogous  to  those  of  (53,  55).     Thus, 

e^  =  cosh  X  +  sinh  x (7); 

e~^  —  cosh  x  -  sinh  x (8). 

Whence,  by  using  the  exponential  series,  we  deduce 

q?  rJ'  ly^ 

sinha;  =  ^  +  ^j+^  +  ^j  + (9); 

cosha;  =  l+|  +  |  +  |  + (10): 

and  from  these  we  may  derive,  by  mere  division,  series  for 
tanh  x^  coth  x,  sech  x  and  csch  x.  The  convergence  of  these 
series  may  be  shown  as  in  the  case  of  the  analogous  series  for 
the  circular  functions. 

64.   Fundamental  Relations. —These,  for  the  hyperbolic 
functions,  are : — 


cosPo;  —  sinh^a;  =1  (1) 

-  sinh  X  .^. 

tanh  a;  =  — -, —  (2) 
cosh  a; 


coshaj 

coth  a;  =  -.— .; —  (d): 

smn£c 

tanh  X  .  coth  x  =  l      (4) 


sinh  X  csch  x     =1     (5); 
cosh  :c  sech  a;      =1     (6); 


tanh^oj  +  sech'^x-  =  1     (7); 

coth^ic-  csch^a?  =  l     (8). 

(2),  (3),  (4),  (5),  (6)  are  evident  from  equations  (l)-(6),  (63). 
To  establish  (1)  we  may  multiply  (7)  by  (8),  (63),  member 
by  member.  To  establish  (7)  and  (8)  we  may  substitute  at 
once  the  exponential  values  of  tanh  x,  etc.,  from  (l)-(G)  (63). 


TRIGONOMETRY.  55 

65.  Limiting  Talues. — Sinhx.  AVhen  ^-  =  0  sinha;  = 
•|(1  -  1)  =  0.  As  a;  increases  sinh  x  rapidly  increases,  becom- 
ing infinite  when  x  —  cc.  If  a;,  diminishing  from  0,  becomes 
negative,  sinh  x  becomes  negative,  and  numerically  increases, 
until,  when  aj  =  -  qo  ,  sinh  a;  =  -  oo  .  It  will  be  observed  that, 
as  in  the  case  of  sin  Xy  changing  the  sign  of  x  changes  the 
sign  of  sinh  x. 

Cosh  X.  When  x  =  0,  cosh  a;  =  1 ,  and  this  is  the  smallest 
value  of  cosh  x;  as  x  increases  cosh  x  increases  very  rapidly, 
until,  when  a;  =  qo  ,  cosh  a:  =  oo  .     Since,  however, 

cosh'-'a;  —  sinh^ic  =  1, 
cosh  X  always  exceeds  sinh  x.     Again,  cosh  x  =  cosh  ( —  x); 
i.  e.,  changing  the  sign  of  x  does  not  affect  the  sign  of  cosh  x. 

Tank  X.  Tanh  0  =^  0,  and  as  x  increases  tanh  x  increases, 
from  0  onward,  but  very  slowly.  As  x  approaches  qo  ,  tanh  x 
approaches  1 ;  when  x  =oo  ,  tanh  a:  =  1.     For, 

e^  _  e--'     1  -  e-^  1  e-^^'  1  1 

=  1,  when  03=  00. 

If  X,  passing  through  0,  becomes  negative,  tanh  x  becomes 
negative,  and  passes  through  the  same  numerical  changes  as  in 
the  case  of  positive  x. 

Coth  X.  AVhen  ic  =  0,  coth  aj  =  l:0=oc;as  x  increases 
coth  X  decreases,  but  remains  the  reciprocal  of  tanh  x,  and, 
when  £c  =  QO  ,  coth  a:  =  1 : 1  =  1. 

For  negative  arcs  the  numerical  changes  in  the  coth  x  are 
the  same  as  in  the  case  of  positive  arcs. 

Seek  X  and  Csch  x.  These  functions  are  the  reciprocals 
respectively  of  the  cosh  x  and  sinh  x.  Hence,  as  x  passes 
from  0  to  oo  ,  sech  x  passes  from  1  to  0,.and  as  x  passes  from  0 
to  —  QO  ,  sech  X  again  passes  from  1  to  0.  As  x  passes  from  0  to 
oo  ,  csch  X  passes  from  cc  to  0,  and  as  x  passes  from  0  to  -  go  , 
csch  X  passes  from  —  oo  to  0. 

66.  Addition  and  Subtraction  Formulae . — By  direct 


ba 


PERIODICITY  OF  THE   FUNCTIONS. 


substitution  of  the  values  (63)  of  the  functions  we  may  readily 

show  that : — 

sinh  (ic  + 1/)  =  sinh  x  cosh  y  +  cosh  x  sinh  2/(1) 
sinh  {x-y)  =  sinh  x  cosh  y  -  cosh  x  sinh  y  (2) : 
cosh  {x  +  y)  =  cosh  x  cosh  y  +  sinh  x  sinh  i/  (3) 
cosh  {x-y)  =  cosh  £c  cosh  y  -  sinh  cc  sinh  y  (4) 
tanh  a:±tanh?/ 


tanh  {x±y')  = 
coth  (a3±t/)  = 


l±tanhaj  tanhi/ 
coth  X  coth  2/±  1 


(5); 


(6). 


coth  2/±coth  X 

We  may  also  derive  these  from  the  corresponding  formulae 
for  circular  functions  (and  vice  versa),  by  using  the  following 
relations,  which  are  evident  from  the  definitions,  viz. : — 

sin  x  =  -  sinh  ix  (1);   tan  x  =  -  tanh  ix  (3) ; 
I  I 

cos  X  =i    cosh  ix  (2)  ;    cot  x  —  i  coth  ix  (4). 

67.  Periodicity  of  tlie  Functions. — The  circular  func- 
tions are  periodic;  i.  e.,  as  x  passes  from  0  to  oo  the  same 
values  of  the  functions  recur  at  intervals  of  2-.  The  functions 
of  ic'are  the  same  as  those  of  {x-\-2n7i),  n  being  any  integer. 
The  values  of  cos  x,  sin  x,  sec  x,  esc  x,  recur  at  intervals  of 
2- ;  the  values  of  tan  x  and  cot  x  at  intervals  of  -.  The  re- 
lations just  expressed  indicate  a  similar  periodicity  in  the 
hyperbolic,  functions.  But  while  the  circular  functions  are 
periodic  for  real  values  of  x,  and  continuous  for  imaginary 
Talues,  the  hyperbolic  functions,  on  the  other  hand,  are  con- 
tinuous for  real  values  of  x  (65),  and  periodic  for  imaginary 
values.     As 

cosh  2nrd  =  cos  "In-  =  1,  and  sinh  2nr:i  =  i sin  2??-  =  0, 
we  have  the  following  formulae,  analogous  in  pairs : — 


CIRCULAR    FUNCTIONS. 

sin  {x  +  2?i-)  =  sin  x ; 
cos  {x  +  2/1-)  =  cos  a; ; 
tan  {x  -H  nrr)  =  tan  x ; 
cot  {x  +   W-)  =  cot  X ; 


HYPERBOLIC     FUNCTIONS. 

sinh  (x  -f-  2n-i)  =  sinh  x ; 
cosh  (x-  4-  27iTd)  =  cosh  x ; 
tanh  (a;  +  nrA)  =  tanh  x ; 
coth  (x  +    7irA)  =  coth  X ; 


TRIGONOMETRY. 


57 


The  student  should  substitute,  successively,  Jri,  -ni,  |ri,  and 
2rt"  for  y  in  all  the  formulae  of  (66),  and  thus  construct  a 
table  for  hyperbolic  functions  analogous  to  that  on  page  14. 

68.  By  processes  analogous  to  those  of  (27,  28)  we  may 
deduce  expressions  for  the  hyperbolic  functions  of  2x  and  of 
^x,  and  also  for  cosh  a^cosh  i,  sinh  a±sinh  b,  etc.  Or,  we 
may  derive  these  formulae  directly  from  the  definitions.     Thus, 

as 

(e^+ib  ^  g-ia-i6>)^gia-i6  ^  g-t«+i6)  =  e«  -f  e"^  +  g"  +  e-\ 

we  have 

cosh  a  +  cosh  6  =  2  cosh  J(a  4-  b)  cosh  ^(a  -  6). 

Similarly  for  the  other  formulae. 

69.  The  exponential  character  of  the  hy}:)erbolic  and  circu- 
lar functions  suggests  important  relations  that  obtain  between 
them  and  the  ordinary  logarithmic  functions.  For  example, 
let  cosh  x^^z,  so  that  sinh  a;=  Vz'^  -  1,  then,  ((7),  (8),  63), 

a;= In  (2;  +  V^z'  -  1),  and  -  x=  In  (z  -  Vz^  -  1). 
Similarly,  if  z  =  sinh  x, 

x^  In  (2  4-  Vz'  +  l),  and  -  re-  In  (l^zN^  -  z). 

*      .      .        l+tanha;     e^        ,_     .^^      •/.        ^     1 
Agam, smce - — - — --  =—,  =  e'^  (6 3 j;  if  «  =  tanh x, 
1  -  tanh  X     e~^ 


2a;  =  In  ^ ,  or  aj  =  ^  In  q- —  =  In-v/  t^ 

1  -z  2      1-z  \  1 


+  z 

-z 


Further,  if  iv=  coth  x,  x-. 


V  1  -  w 


The  corresponding  relations  for  the  circular  functions  are 
evidently, 

xi:=  In (Vl^'  +  zi)=  -  In (VY^'-zi)  (1); 

70.  This  connection  of  the  Trigonometric  functions  with 
logarithms  indicates  that  we  are  not  confined  to  the  natural 
base,  e,  in  our  definitions.     Thus,  if  a  be  the  base,  instead  of  e, 


58   GEOMETRIC  INTERPRETATION  OF  HYPERBOLIC  FUNCTIONS. 

since  a^=e^^°«,  if,  in  all  discussions  we  replace  a  bye,  the 
only  change  required  in  the  functions  will  be  the  change  of  the 
exponent  from  x  to  x  In  a,  which  will 
evidently  not  affect  the  nature  or  the 
results  of  our  investigations. 

71.  Geometric  Interpretation 
of  the  Hyperbolic  Functions. — In 
(4)  we  dealt  with  the  relations  of  the 
Fig.  23.  circular  functions  to  the  sides  of  tri- 

angles of  reference.  Let  us  now  investigate  the  analogous  re- 
lations of  the  hyperbolic  functions.  If  BAG,  (Fig.  23),  be 
right-angled  at  A,  we  have 

sin  C  =  c :  a ;  cos  C  =  6  :  a ;  tan  C  =  c:h. 
Kow  let  BA  revolve  about  B,  the  angle  C  remaining  con- 
stant, and  let  x  be  an  arc,  some  function  of  the  measuring  arc 
of  C,  such  that 

sinha;  =  c:a;  cosha;=6:a;  tanhic  =  c:6. 
We  are  to  determine  the  proper  shape  of  the  triangle,  and 
the  relation  betwaen  C  and  x.     Since,  (22), 
sin  C     c^  sin  B     6,  sin  C     c 
sin  A     a   sinA~a   sinB     6' 
and  since  cosh^a;  -  sinh^aj  =  1  (or);  we  obtain,  on  substituting  in 
(or)  for  cosh  X  and  sinh  x  their  values, 

sin'B  -  sin'C  =  sin^^A  =  sin"'  (C  +  A)  -  sin'C ; 
or,  since  sinV  -  sin^^/  =  sin  (x  +  y)  sin  (x  +  y) 
sin  (2C  +  A)  sin  A  =  sin^A;  whence 
sin  (2C  +  A)  =  sin  A  (1),  or  sin  A  =  0  (2). 
The  latter  value  of  sin  A  being  inadmissible,  we  have,  from  (1), 
2C  +  A  =  --A,  orC  +  A  =  j7:; 
whence  B  =  ^r;  L  e.,  BA  ±  CB. 
This  determines  the  shape  of  the  triangle.     The  relation  be- 
tween C  and  X  is  evident;  for 

sinh  a?  =  tan  C ;  tanh  a?  =  sin  C; 
cosh  a;  =  sec  C ;   coth  x  =  esc  C. 
72.  If,  in  any  circle,  (Fig.  24),  a  perpendicular,  PM,  be 
dropped  from  the  termination,  (M),  of  any  arc  on  the  diame- 


TRIGONOMETRY. 


59 


ter  through  the  origin  thereof,  and  if  PM  =  y,  OM  ic,  and 
radius  =  a,  then  evidently, 

Hence  we  may  at  once  write : — 

ic  =  a  cos  ^;  y  =  a  sin  ^,  if  6  denote  the  radial  measure  of 
AOP.  This  convention  gives 
us  the  circular  functions  as  de- 
fined geometrically  in  Chapter 
I.  \a^d  is  the  area  of  the  sector 
AOP,  and  hence,  if  a  =  1,  d 
represents  double  this  area. 

The  hyperbolic  functions  are 
connected  in  an  analogous  way 
with  the  rectangular  hyperbola. 
Thus,  (Fig.  25),  if  PM  be  the 
perpendicular  dropped  from  any 
point  P,  of  a  rectangular  hyper-  ^*^'  ^^' 

bola  on  the  diameter  through  the  origin,  (A),  of  the  hyperbolic 
arc,  and  if,  as  before,  PM  =  y,  OM  =  x,  and  the  radius  of  the 
auxiliary  circle  be  a.  we  have,  from  the  well-known  property 
of  the  curve. 


Hence  we  may  define  the  hyperbolic  functions  geometrically 
as  follows: — 

(1)  The  hyperbolic  sine  is  the  perpendicular  dropped  from 
the  termination  of  the  hyperbolic  arc  on  the  diameter  through 
the  origin,  divided  by  the  radius  of  the  auxiliary  circle. 
This  radius  is  to  the  hyperbola  what  the  radius  is  to  the  circle, 
viz.,  the  semi-axis  of  the  curve. 

(2)  The  hyperbolic  cosine  is  the  distance  from  the  centre  to 
the  foot  of  the  hyperbolic  sine  (the  base  of  the  hyperbolic 
triangle  of  reference),  divided  by  the  semi-axis  (a). 

In  a  circle  every  diameter  is  an  axis;  hence  any  semi-diameter  (or  radius)  may  be 
taken  as  a  semi-axis.  In  the  hyperbola,  however,"  tliis  is  not  the  fact;  hence  we  take 
the  only  semi-diameter  that  is  also  a  semi-axis,  as  the  denominator  of  the  hyperbolic 
functions. 


60    GEOMETRIC  INTERPRETATION  OF  HYPERBOLIC  FUNCTIONS. 

The  definitions  of  the  geometric  representatives  of  the  other 
hyperbolic  functions  are  at  once  evident  (63). 

73.  As  will  be  shown  later,  the  area  of  the  hyperbolic 
sector  AOP  is  ^a\  (u  being  the  hyperbolic  arc  AP),  and 
hence,  just  as  in  the  case  of  the  circle,  if  a  =  1,  i*  is  double  the 
area  of  AOP. 

From  M  draw  ML,  tangent  to  the  auxiliary  circle  (of  radius 
a),  at  L,  and  denote  the  angle  (arc)  AOL  by  6.     Then 
a  sec  ^  =  OM  =  a  cosh  u ; 
a  tan  ^  =  LM  =  a  sinh  u  =  PM. 

When  6  is  thus  connected  with  u  it 
is  called  the  Gudermannian  of  t*,* 
ordinarily  abbreviated  gd  u.  A  tri- 
angle of  reference  for  any  hyperbolic 
function  of  u  may  be  readily  con- 
structed ;  the  calculation  of  any  hy- 
perbolic function  may  thus  be  reduced 
to  the  calculation  of  a  circular  func- 
Fig.  25.  tion.     For  example,  lay  off 

MK  =  a,  and  draw  PK ;  tan  MKP  =  PM :  a  =  sinh  u. 
Similarly  for  the  other  functions. 

74.  Hyperbolic  and  circular  functions  are  of  great  impor- 
tance in  the  Integral  Calculus,  where  it  will  be  shown  that  they 
are  special  forms  of  certain  more  general  functions,  known  as 
Elliptic  Functions.  Any  attempt  at  a  discussion  of  these  latter 
would,  however,  be  entirely  beyond  the  scope  of  this  work. 

Interesting  examples  of  the  methods  of  employing  these 
functions  in  the  solution  of  equations  will  be  found  in  Tod- 
hunter's  and  Burnside  and  Panton's  treatises  on  the  Theory  of 
Equations,  and  also  in  the  various  mathematical  journals, 
among  which  Crelle's  is  perhaps  the  best.  These  discussions 
are  not,  however,  deemed  of  sufficient  practical  importance  to 
warrant  their  introduction  into  the  present  work. 

*cf .  Orelle,  "  Journal  fUr  die  reine  und  an^'ewandte  Mathematik,"  Band  6. 


EXERCISES- 


CHAPTER  L 

1.  Construct 

sin-Y.3);  tan-^e^;  cof^C  -  4);  sec-'(  -  3);  coS"\  -  .  6); 

sin~'j  -  |;  tan~^7r;  csc~V  —1  |;  cos~M  -  J. 

Determine  (with  a  protractor)  these  angles,  to  within  1°; 
I       determine  also  their  radial  measure. 

2.  Determine,  with  protractor  and  scale,  the  functions  of 
5°,  10°,  15°,  etc.,  to  180°  inclusive. 

3.  Determine  geometrically  the  functions  of  (90°  ±a;), 
(180°±a:),  (270°tx),  and  (360°±ic),  in  terms  of  functions 
of  X. 

4.  Find  the  functions  of 

sin-^'-  ;   cos"'^ ;  tan-^c ;  cot-^(  -  2);  csc~\  -  m). 
5  0 

5.  Express  successively  all  the  functions  of  x  in  terms  of 
each. 

6.  State  the  signs  of  the  functions  of 

(2?i  +  l)180°±A;  2?ir±A;  (2?i.  -  l)r- ^^^  rr. 

7.  Prove  that 
sec  X  +  CSC  X     1  +  cot  X     tan  ^  + 1 


(1) 


secic  — cscx     1-cota;     tanic-1 


.^^    sec  X  +  CSC  X  -  sec  x  esc  x  sin  a:  cos  x  sec  x  esc  x 


sec  X  CSC  X  sec  x  +  esc  x  +  sec  x  esc  x 


.^^         1  H-tan'a; 

(^)    A/r" -^=tanic;   tan  a;  +  cot  x  =  ^  sec  a;  esc  a;. 

^        \  1+COt'iC 

^]:r^<^-  ^  (1) 


EXERCISES. 


(4)  — -  =  - ^ r— .tan^aj. 

CSC  a;  -  1     1  +  sina;    coversm  x 

(5)  tan-^^  +  cot-X2aJ  +  l)  =  ^ 

(6)  tan  A  +  cot  A  =  sec  A  esc  A. 

1 


(7)   secB  +  V'sec=^B-l  = 


sec  B  -  -y/sec^B  - 1 


(8)  ]     'l^'f  =  l  +  2tan^A-2secAtaiiA. 
1  +  sin  A 

.(.-.    tan  A  +  tan  B     ,       .  ^     _, 

(9) =:=tanAtanB. 

^       cot  A  +  cot  B 

(10)   sec-^Acsc'A-(sec'A  +  csc'A)  =  0. 


(11)   sinA  +  cosA=     /?5^A^!:^A±^. 
\      sec  A  CSC  A 

8.  Solve,  analytically  and  graphically,  the  following  equa- 
tions : — 

(1)    2  cos'^a;  =  tan  x ;    (2)  esc  a;  =  4  sin  x ;   (3)    sin  cc  =  2  cos  x, 
(4)   sin  a;  +  cos  ic  =  CSC  ic ;  (5)  5  sec'^ic  — 12  sin^a;  =  4  tan  x. 

(6)  sin  X  cos  a;  +  tan  a;  cot  x  = . 

sec  a;  esc  x 

(7)  2  (sin  X  +  cos  a;)  +  sec  a;  esc  a;  -  cot  x  =  S. 

(8)  3  sec  X  esc  x  tan  x  =  cot^a;  + 1 . 

(9)  3sina;  +  5cosa;  =  8;  (10)  2  tan  aj  +  3cscaj  =  6. 
(11)  cot  aj  -  4  sin  x  --=.7  ;  (12)  sin-'a;  +  7  cos  a;  =  1. 

(13)  3  cos^a;  +  6  sin^a;  cos  x=  1. 

(14)  secaj— tanaj=  4'v/sina;-2. 

(15)  a  cos  x  +  b  sin  x=  c. 

9.  Show  that,  for  any  angle, 

2  (sin*A  +  cos*A  +  2  sin^A  cos'^A)  = 

(tan  A  +  cot  A)^  -  tan^A  —  cot''^A=  constant  ; 

cos  A  +  COS  B      sin  A  +  sin  B 

sin  A  -  sin  B     cos  B  -  cos  A 

10.  Solve  the  equations      r  sin  x=a  (1) 

r  cos  a; =6  (2), 


CHAPTER   II. 

1.  Solve  the  right  triangle  ABC,  given 

(1)  6=    3;c=4  ;   (4)   a=625      B^52°13'; 

(2)  a=29;6=20  ;   (5)    6=.00216  B==  87°26'; 

(3)  a=  30.126  ;  C=27°  50' ;  (6)   c=^. 00326  C=:    2°  15'. 

2.  When  the  sun's  altitude  is  36°  19'  the  shadow  of  a  flag, 
pole  is  175  ft.  What  is  the  height  of  the  pole?  What  is  the 
length  of  the  shadow  when  the  sun's  altitude  increases  to 
40°  16' 27"? 

3.  A  winding  road  3  miles  long  has  a  rise  of  1,196  feet. 
Required  the  angle  of  (uniform)  slope. 

4.  The  angles  of  elevation  of  the  top  of  an  inaccessible  crag 
are  32°  46',  and  24°  17',  as  observed  from  two  points,  200 
yards  apart,  in  the  same  vertical  plane  with  the  top.  If  the 
theodolite  stands  four  feet  from  the  ground,  what  is  the  height 
of  the  crag? 

5.  Two  towers  stand  on  opposite  banks  of  a  river.  From 
the  top  of  one,  168  feet  high,  the  angles  of  depression  of  the 
top  and  bottom  of  the  other  are  observed  to  be  13°  28',  and 
28°  16',  respectively.  Required  the  breadth  of  the  river  and 
the  height  of  the  second  tower. 

6.  What  is  the  angle  of  depression  of  a  tangent  to  the 
earth's  surface  as  observed  from  the  top  of  a  mountain  18,260 
feet  high  ?  What  is  the  distance  from  the  foot  of  the  mountain 
at  which  the  top  must  cease  to  be  visible,  the  average  slope 

being  28°  17'? 

7.  Let  E  represent 

the  earth  and  M  the 

moon.   The  angle  EMS 

=  57',  (moon's   paral- 

Fij:.  i6.  lax);  the  radius  of  the 

earthis  3,956  miles.   Required  thegeocentric  distance  of  the  moon. 

8.  The  angle  MET  (the  moon's  apparent  semi-diameter),  is 

31'  20"      What  is  the  diameter  of  the  moon  ? 

(3) 


4  EXERCISES. 

9.  To  find  the  point  at  which  a  perpendicular  from  an  in- 
accessible point  will  meet  a  given  line. 

10.  In  a  right  triangle  are  given  R  and  r.     To  solve. 

11.  Find  the  volume  of  a  cone  of  revolution  whose  vertical 
angle  is  75°  16'  23",  and  whose  slant  height  is  2a  136. 

1 2.  To  solve  an  isosceles  triangle,  when  the  base  and  r  are 
given. 

13.  Each  side  of  the  base  of  a  regular  heptagonal  pyramid 
is  10.307,  each  lateral  edge  makes  with  the  base  an  angle  of 
39°  46'  19".     Required  the  volume  and  surface  of  the  pyramid. 

14.  To  solve  a  right  triangle,  given  R  and  B,  or  R  and  h. 


CHAPTER  HI. 


1.  Establish  geometrically  the  formulae  of  (27-8-9). 

2.  Show  that:— 

(1)  sin  (30°  +  A)  +  sin  (30°  -  A)  =cos  (60°  +  A)cos(60°  -  A) 

=  cos  A ; 

x^x        -i  A     cotiA  +  tanA  .     cotiA  +  taniA 

(2)  sec4A= ^-— r-i ;  secA=— -f-i — r — ?-*-: 

^  ^       ^  cotJA  cot^A-tan^A' 

(3)  sin  3a7=  sin  x  (4  cos'^cc— 1)  =  3  sin  a;  -  4  sin^ 
sin  4ic  =  sin  £c  (  .8  qos^x  -      4  cos  .^•), 

sin  bx=  sin  x  {  \%  cos*a;  -    12  co^^x  + 1), 

sin  6ic= sin  x  (  32  cos^:^  -    32  cos^an-   6  cos  a;), 

sin  7a;  =  sin  a;  (  64  cos^'a;  -    80  cos*a;  +  24  cosmic  —  1), 

sin  80;= sin  x  (128  cos'a;-  192  cos^it'H-  80  cos^a?-8cosa;); 


(4)  cos  3a;  =     4  cos^a;  -      3  cos  x, 

cos  4a:  =     8cos*a;-      8cos^a;+      1, 

cos  bx^  16  cos^a;  -    20  cos*.r  +      5  cos  x, 

cos  Qx=  32  cos^a;  -    48  cos*a;  +    18  cos^a;  -    1, 

cos  7a;=  64  cos'a;  -  1 1 2  cos^'c  +    56  cos^a;  -    7  cos  x, 

cos  8a;- 128  cos^a;  -  256  cos^a;  +  160  cos*a;  -  32  coslr  + 1. 


EXERCISES.  5 

Establish  also  the  formulae  for  the  tangents  and  cotangents 
of  multiples  of  x, 

Gos^k=^-^  cos  2^  +  —  . 

1  3 

cos^^=  -J-  cos  3/:  +  —  cos  h, 

cos*^=  -r-  COS  4^  +  -^  cos  2^  +  -^ , 

0  L  o 

C0S^^=-Tr77   COS  ^h-V—t  COS  3^  +  -5-  COS  h, 

lb  lb  o 

C0S^^=-:r7r   COS  6^  +  :j7;  COS  4^  +  r^  COS  2^  +  :^  , 

oz  lb  Iz  lb 

cos^^=-rr  cos  Ih  +  7,-:  cos  5^  +  ^  :  cos  3^  +  tt.  cos  h, 
b4  b4  b4  b4 

1  1  7  7  35 

-    C0S^^=r7T^  cos  %h  +  —  COS  GA;  +  — -  COS  4^  +  77^  COS  'Ih  +  zrr^--  . 

Izo  lb  oZ  lb  Izo 

3.  Show  that : — 

tan.x±tanw=  — ^-^^  (1);  tana;  +  cotw= ^-t^^-          (3j; 

cosxcosy  cos^smi/ 

^     ,      ^        sin  (w+x)    ,^^  /      ,     X     l+tanrt-tan^ 

cota;±cotw=^ — ^T       (2);    cos(a;±  w)  =  --:- ^  (4^ 

sm  x  sin  1/  ^        ^  ^        sec  xaecy      ^  ^* 

What  do   (1),  (2),  (3),  become  when 

x  =  a-^-,  and  ?/=  a  + 1-- ? 

(5)  sin  (aj  +  y)  sin  {x-y)-=  sin^cc  -  sin"^^/^  cos^i/  -  cos'^x ; 

(6)  cos  (aj  +  2/)  cos  {x  -y)  =  cos'^a^  -  siii^y  =  cos^^/  ~  sin^^  • 
From  (6)  deduce  the  value  of  cos  2x. 

(7)  (tan'a:  -  tsLii'y){cofx  -  coi^y)=  ? 

J     au        ^u  ^  ^      1        1-cosaj        ,      ^,        1  +  cosa; 

4.  fenow  that  tanj.7;=  — : ,  and  cot  Jo;  = 


smic 


Prove  that  tan2ic=  — ^ — ;  also  that 

COSiC  +  COSdx' 

^       sin  2x  +  sin  4x 
tan  3a;  = 


tan  4x  = 


cos  2x  +  COS  4x ' 

sin  X  +  sin  3a;  +  sin  5a;  +  sin  7 


cos  X  +  cos  3a;  +  cos  ox  +  cos  7a; 


X 


tan  5x  = 


EXERCISES. 

sin  2x  -h  bin  4x  +  sin  Qx  +  sin  8a: 


cos  2x  4-  cos  4£P  +  cos  6x  4-  cos  Sx ' 
Deduce  the  corresponding  expression  for  tan  nx. 
Show  that  cot  x=  tan  x  +  2  tan  20?  +  4  tan  4ir  +  8  tan  Sx. 

5.  Prove  that : — 

(1 )  cos  (x  +  y)  +  sin  (x  -  2/)  =  -/  2  sin  ( Jrr  +  x)  cos  ( J-r  ±  ?/) ; 

(2)  cos  i/=  cos  (x  +  2/)  cos  ;c  +  sin  {x  ■\-y)smx; 

(3)  tan-^^  +  cot-V2a;  + 1)=  7- ; 
^  ^  ic  +  1  4 

(4)  2cot-'3  +  cot-7  =-r; 

/rx  .       1       .       1       .       A-x-y-xy     TT 

(5)  tan-U-  +  tan-V  +  tan-^^; ^^ =  7  • 

^^  ^  l+x  +  y-\-xy4 

6.  Given  sin  2.r  =  cos  Sx ;  find  x  and  its  functions. 

7.  From  the  functions  of  30°,  45°,  and  18°,  find  the  func- 
tions of  all  those  angles  between  0°  and  90°,  the  expressions  for 
which  involve  no  fraction  less  than  0.°5,  and  the  calculation 
of  whose  functions  requires  no  other  data. 

8.  Show  that:— 

W.    ^        2  tan  x                2               r.     r^  cos^a-  +  sin'ic 
sin  2x  = ~  = =  2-2  : — : 
1  +  tan-a;     tan  x  +  cot  x              cos  x  +  sin  x 

cot'a;-!  ^  .  cot  ccipl 

(2)    cos  2x=  -— ~  =  (l±sin2a:) -V  ; 

^  ^  cot'a3  +  l       ^  ^cot^±l 

X       1+sin^  +  cosiP 

,     (3)     cot  -.     =  :; ; ^ 

\  2       1  +  sin  0?  -  cos  a; 

i   9.  Express,  as  the  sum  or  difference  of  two  functions, 

(1),  2  sin  3a;  cos  2  2/;  (2)  sin  3.r  cos  5a; ;  (3)  sin  20°  sin  30°. 

^.      -.„     sin  3a;±sin  5a;     cos  4;i;±cos 6.^' 

bimpliiy =— ;  — — : — -— . 

cos  3a;±cos  5a;      sm  4a;+sni  bx 

Prove   that   tan  5a;  -  tan  3a;  -  tan  2a;  =  tan  5x  tan  3a;  tan  2a;. 

1 0 .  Solve  the  following  equations : — 

(1)  cos  X  -  sin  X  =  (cos  x  +  sin  a;)(2  cos^a;  -  1); 

(2)  cos  (a;  +  -5-)  +  2  sin  a;  +  3  tan  (x  +  —-)  =  cot'-^a;  +  esc  x ; 

(3)  2  tan  a;  +  tan  (a- a;)  =  tan  (6  4- a;); 


EXERCISES. 


(4).    2a  sec  X  =  (a  -\-  b^O-  -  tan  x  +  sec  x) ; 

(5).  ^/smx  -  CSC  a;  -  l/ 1  -  esc  a;  =  esc  x  (sin  x  -  1). 

11.  Show  that  :— 

sin  A  +  sinB  +  sin  (A  +  B)  A        B  ... 

r— :5 ^-)-r ^    =COt^  cot  -  (1): 

sin  A  +  sin  B  -  sm  ( A  +  B)  2         2  ^  ^ 

tan  3A  cot  A  (2  cos  2A  - 1)  =  2  cos  2A  + 1  (2) 

12.  Sum  the  series,  sinic  +  asin2ic  +  rt'^sin  Sx  + 

Assuming  that  a?  =  cos  m  +  V  -  2  sin  m,  sum  x  +  x'^  +  x^  +  ,,  . 

13.  Show  that,  if  C  =  90°, 


a  +  h     a  —  b     4  sin^A 


-2. 


1 

COSOJ 

COS  2/ 

0 

COS  a; 

1 

cos  2 

0 

cosy 

COS  2! 

1 

COS  w 

0 

0 

COS  11^ 

1 

a  -  6     a  +  6      COS  2B 

14.  COS  (x +  y  +  z)  + cos  (^x  +  y  —  z)  +  cos  (^x— y -^  z)  + 

_  cos(z  +  y-x')  =  ^  y__ 

15.  Show  that  if  ^^^^^^tkavv-ao^. 


=  0, 


sin'''a5  cos'^w;  =  1  -  cost's  -  cos'-'y  ~  cos'^o;  +  2  cos  ic  cos  y  cos  z. 

16.  Evaluate: — 

1,  1,  1,  since,  sin 2/,  sin 2, 
sin  aj,  sin  y,  sin  2,  cos  a;,  cosy,  cos  2, 
cos  x,  cos  y,  cos  2,    ,   sin  2a;,  sin  2y,  sin  22, 

1 7.  Given,  (1),  sin  £c  -  J  sin  2x  =  m; 

(2),  cos x-^  cos 2a;  =  ^ :  find  a;. 
18'  Eliminate  x  and  1/  from  the  groups  of  equations: — 
r  cos  X  sec  y--  (r:  a)  esc  (a;  -f  y) 
=  [(t :  6)  -  sin  x  sec  y]  sec  (^x  +  y)  =  tan  3/  (1)^ 

r  cos  (x  — 37/)  =  2a  cos^i/ 5   ^  sin  (a;  —  3?/)  =  2a  sin^y    C^)- 
19-  Show  that  two  geometrical  progressions  may  be  formed 

firom  the  roots  of 

x^  -  2x  cos  a  + 1  =  0, 
x'-  -  2x  cos  2a  + 1  =  0, 


sin  x,    sin  ?/,    sin  2, 
sin  2x,  sin  2y,  sin  22, 
sin  4a;,  sin  4y,  sin  42, 


or*  -  2a;  cos  Tia  +  1  =  0; 


EXERCISES. 


the  product  of  the  two  series  being  sin^  I  —  jcsc^  (  o  I 
20.  Show  that:— 


a" 
6  sin  A 
c  sin  A 

0 
h 
c 

21. 


6  sin  A 
1 

cos  A 
h 

1 
cos  A 
Show  that 

cosi  (B-C) 
cos  4  (B  +  C) 
sin  4  (B  +  C) 


c  sin  A 

1 

cos  A 

=0; 

1 

c 

COS  A 

- 

6- 

1 

e- 

-1 

cosC 

cosC 

-1 

cosB 

cosA 

a 

h 

ccosA 

a 

h  cos  A 

0 

cosB 

cos  A 
-1 


0; 


=  a^ 


equals 

22, 


cosi  (C-A)  cos^  (A-B) 
cos|(C  +  A)  cos|(A  +  B) 
sinJ(C  +  A)     sin|  (A  +  B) 

sin  (B  -  C)  +  sin  (C  -  A)  +  sin  (A  -  B). 
If  A,  B,  C,  be  the  angles  of  a  triangle : — 
cos  A  cos  B  cos  C  ^ 

-: h  -: ; h  -; : '  =  2 

sin  B  sin  C     sin  A  sin  C     sin  B  sin  C 
sin^C  =  cos"B  +  cos'^A  +  2  cos  A  cos  B  cos  C 
sin  A  +  sin  B  +  sin  C  =  (semi-perimeter)  :  R 
e+h:a::e-h:c  cos  B  -  5  cos  C 
&  cos  (  X  -  C)  +  c  cos  (X  +  B)  =  6  cos  (X  +  C)  +  c  cos  (X  ■ 
he  +  4R^  =  cos  A  +  cos  B  cos  C  =  2  (Area)  sin  A  +  a^ 
2R  sin  C  =  6  cos  A  +  \/{a'  —  b'^  sin'^A) 
23.  What  is  the  shape  of  a  triangle  in  which 

B)? 


(1); 

(2); 

(3); 

(4); 

B)(5); 

(6); 
(7). 


a'  +  h' :  ce  -  h'  =  sin  ( A  +  B) :  sin  (A 

24.  In  any  right  triangle,  (A  -  90°):  - 

sin  2B  +  sin  2C  =  4  a^  +  c^  (1); 

sin  2B  -  sin  2C  =  cos  2B  +  cos  2C  =  0  (2); 

sin^C  +  sin^B  -  (sin'^JB  +  sin'^C)  =  {h  +  c>+  2a  (3). 

Express  (3)  in  terms  of  cosines. 

25.  In  any  triangle,  ABC, 

6  -  c  :  a  : :  sin  ^(B  -  C)  :  cos  ^A  : :  cos  C  -  cos  B :  1  +  cos  A; 
6  +  c  :  a  : :  cos  J(B  -  C)  :  sin  J  A : :  cos  C  +  cos  B  :  1  -  cos  A; 
a  +  i  -I-  c  =  (6  +  c)  cos  A  +  (c  +  a)  cos  B  +  (a  +  6)  cos  C  ; 
(cos  A  -f  6  sin  C)  +  (cos  B  +  c  sin  A)  +  (cos  C  -r  a  sin  B)  = 
2  (sinA  +  2sinB)-r-(a  +  2i»). 


EXERCISES.  9 

26.  If  i,j,  h  be  the  altitudes  of  a  triangle, 

a  sin  A  +  i  sin  B  +  c sin  C  =  2(icos  A  +j  cos  B  +  ^cos  C)         (1); 
sinA  =  sin(2A4-B  +  C)  =  -cosK3A  +  B  +  C)  (2); 

tan  (B  -  2C)  =  -  tan  (A  +  2C)  (3); 

csc(P-C)  =  sec(iA  +  C);tan(A  +  2B  +  3C)=tan(A-C)(4); 
cot  ^(B  +  C)  cot  ^(C  +  A)  -f  cot  J(C  +  A)  cot  i(A  +  B) 

+  cot  |(B  +  C)  cot  ^(C  +  A)  =  1  (5). 

27.  Extend  to  quadrilaterals  (and  to  polygons  in  general), 
the  relation  a^=  fe^  +  c'^  — 26c  cos  A.     Extend  also  the  relation 

a  =  b  cos  C  +  c  cos  B. 

28.  The  roots  of  a  cubic  are  the  tangents  of  the  angles  of 
a  triangle.  Required  the  form  of  the  cubic.  Investigate  also 
the  cubic  whose  roots  are  sines  and  cosines. 


EXERCISES  IN  SOLVING  PLANE  TRIANGLES. 

1.  To  solve  a  triangle  ABC,  given: — 

(1)  a  =  237.062,  h  =  127.268,  A  =  27°  32'  16". 

(2)  a  =  156.82.    6  =  284.072,   c=  192.007. 

2.  To  solve  a  triangle  ABC,  given: — 


_  (1)    a,  6,A±B 

U^(2)  A,a,6^±c^ 

(3)  A,  C,  a±6: 

Jk-  (4)   c,A,a'^±6^: 


(5)  The  altitudes; 

(6)  The  medians ; 

(7)  The  bisectrices  of  the  angles ;      ^^ 

(8)  The  sides  of  the  Pedal  triangle ; 
(9)  The  perimeter,  an  altitude  and  a  side ; 

(10)  The  radius  of  the  inscribed  circle,  a±h,  and  c; 

(11)  The  radius  of  the  inscribed  circle,  the  radius  of  the 
circumscribed  circle,  and  one  altitude. 

3.  The  distance  between  two  buoys,  A  and  B,  being  required, 
a  base  line,  CD,  of  300  yards  is  laid  oif,  and  the  angles,  CDA, 
(  =  28°  19'),  ADB(  =  62°16'),  ACB  (  =  57°38'),  and  BCD 
( =  38°  52')  are  measured.     Required,  AB,  and  the  angle  CBD. 

4*.  Deduce  expressions  for  the  side,  apothem  and  area  of 
each  regular  polygon  inscribed  in  a  circle  whose  radius  is  10, 
up  to  and  including,  the  dodecagon. 


10  EXERCISES. 

5.  An  aeronaut  observes  the  angles  of  depression  of  two 
towers  on  opposite  shores  of  a  lake,  and  12  miles  apart,  to  be 
47°  58'  and  33°  25',  respectively.     What  is  his  altitude? 

6.  An  engine  passes  a  station,  C,  that  is  8  miles  distant, 
and  that  beai-s  S  32°  17'  W,  at  10  A.  m.;  moving  continuously, 
it  passes  B,  N  68°  26'  W,  and  5  miles  away,  at  10 :  20  a.  m.  If 
the  road  BC  is  straight,  what  is  the  average  rate  of  speed  of 
the  engine  ? 

«     n,         1    ,    /T^-     rf   ON    ^-r^      c  sin  A  sin  B 

7.  Show  that,  (Fig.  7,  8),  CD= — ^— ^- — ^— ;  express  the 

area  of  ABC  in  terms  of  c.  A,  B. 

8.  The  area  of  the  largest  circle  that  can  be  cut  from  a 
triangular  block,  is  1312.  The  area  of  the  block  is  2018,  and 
the  angle  at  one  corner  is  57°  10'.  Required  the  dimensions  of 
the  block. 

9.  If  ABC,  DEF,  have  a  =  d,b  =  e,  and  C  =  180°-  F,  they 
are  equivalent.     Generalize  this  statement. 

10.  To  determine  the  distance  from  D  to  an  invisible,  inac- 
cessible point,  B. 

11.  Show  that  in  any  triangle,  ABC : — 
cot^A-rCot  JB  +  cot|C  =  s-^r; 

sin  A  +  sin  B +  sin  C  =  s-^  R  =  sin  A  sin  B  sin  C  •  2  R-7-r ; 
tan  A  +  tan  B+tan  C  =^  tan  A  tan  B  tan  C ; 
cos^A  +  cos^B  +  cos'^C  =  1  -  2  cos  A  cos  B  cos  C. 

12.  A  line  AD  is  drawn  from  A  to  BC.  Compare  the 
ratios  D AC :  DAB,  and  DB  :  DC. 

1 3.  To  find  the  area  of  an  inaccessible  field,  knowing  a  base 
line  KL,  and  the  angles  subtended  at  K  and  L  by  the  sides  of 
the  field. 

17.  Express  the  area  of  a  quadrilateral  in  terms  of  the 
diagonals  and  their  included  angle. 

18.  Determine  a  trapezoid,  knowing  the  sides  a,  b,  c,  d;  a 
being  parallel  to  c. 

19.  B  is  due  south  of  A.  An  observer  at  C  notes  the  re- 
spective intervals  (6  seconds,. 9  seconds),  between  the  flashes  of 


EXERCISES.  11 

the  sunset  guns  at  B  and  A,  and  the  times  at  which  the  reports 
reach  him.  AB  subtends  to  him  an  angle  of  68°  29'.  How 
far  is  A  from  B  ? 

20.  A  circular  arch,  98  feet  in  (curvilinear)  length,  rests  on 
piers  16  feet  high  and  82  feet  apart  (from  centre  to  centre). 

,What  is  the  height  of  the  highest  point  of  the  arch? 

21.  Three  circles,  of  radii  8,  12,  and  20,  respectively,  are 
tangent,  each  to  the  other  two.  Required  the  perimeter  and 
area  of  the  enclosed  space. 

22.  Find  the  distance  in  space  described  in  one  day,  in  con- 
sequence of  the  rotation  of  the  earth,  by  the  Great  Equatorial 
at  the  Lick  Observatory  of  the  University  of  California,  at  ^It. 
Hamilton,  (Lat.  37°  20'  22".51  K     Long.  121°  38'  35".75  W). 

23.  Find  the  distance  between  the  orthocentre  and  the  cen- 
tre of  the  circumscribed  circle,  (of  ABC),  in  terms  of  the 
radius  of  the  latter  and  the  angles  of  ABC. 

24.  Express,  in  terms  of  the  radii,  the  distance  between  the 
centres  of  the  inscribed  and  circumscribed  circles.  Express, 
(in  terms  of  a,  h,  c),  the  distances  between  the  centres  of  the 
escribed  circles. 

25.  Show  that  the  area  of  ABC  =  4Rr  cos  |A  cos  JB  cos  |C. 
20.    Circles  passing  through  the  centre  of  the  circumscribed 

circle,  and  having  a,  h,  c  as  chords,  have  for  radii  k,  I,  m). 
Prove  that  {(i^k)  +  {h^l)  +  {c-^  m)  =  8  sin  A  sin  B  sin  C. 

27.  Show  how  to  determine  the  height  of  a  cloud  from  its 
angle  of  elevation  and  the  angle  of  depression  of  its  reflection 
in  a  lake  h  feet  below  the  level  of  the  observer. 

28.  A  man  sees  two  spires,  the  tops  of  which  seem  to  be  in 
a  straight  line  whose  angular  elevation  is  m ;  the  angles  of  de- 
pression of  the  reflections  in  a  lake  h  feet  below  him  are  B  and 
C.  Determine  the  distance  between  the  spires,  and  the  height 
of  each. 

29.  In  any  triangle,  E  = 

J(a  cot  A  CSC  B  CSC  C+  6  cot  B  esc  A  esc  C+c  cot  C  esc  B  esc  A). 


12  •  EXERCISES. 

BO.  Determine  the  radii  (/,/',/'' )of  the-escribed  circles,  in 
terms  of  the  sides.     Hence  show  that 

-  +  —+-777  = (1);   area=-/rrVV"    (2); 

r      r'     r        area  ^  ' 

/  +  /'  +r'"-r  =  4i^(3). 

31.  A,  walking  along  a  straight  road,  observes  that  the 
greatest  angle  that  two  distant  spires  subtend  at  his  eye  is  a\ 
walking  c  rods  farther  he  observes  that  they  are  in  a  line  mak- 
ing an  angle  ft  with  the  road.  Find  the  distance  between  the 
spires. 

32.  A  spire  S  is  observed  from  A  and  B.  If  the  angles 
SAB  and  SBA,  the  distance  AB,  and  the  angle  of  elevation  of 
the  spire  at  B  be  known,  find  the  distance  of  the  spire  from  A 
and  its  height  above  the  level  road  AB. 

33-  The  angular  elevation  of  a  hill,  as  observed  from  a 
point  due  north  of  it,  is  A;  as  observed  from  a  point  due  west 
of  the  first  and  m  miles  distant  across  a  lake,  the  angle  of  ele- 
vation is  B.     How  high  is  the  hill  ? 

34.  An  observer  at  sea  takes  the  angles  subtended  at  his 
position,  (P),  by  the  known  shore  distances,  AB,  AD,  DB. 
Find  AP,  DP,  BP,  when  APD  =  42°  17',  BPD  =  36°  25',  AD  = 
2100,  BD  =  1672,  AB  =  2867. 

35.  To  determine  the  geocentric  distance  of  a  heavenly 
body  M,  knowing  the  zenith  distances  of  M  at  two  stations  of 
which  the  latitudes  are  known,  and  which  are  situated  on  the 
same  meridian. 

36.  To  determine  the  area  of  the  greatest  pentagonal  star 
that  can  be  cut  from  a  circular  disc,  four  inches  in  diameter. 

37 .  The  area  of  the  base  of  a  regular  quadrangular  pyra- 
mid is  289  square  feet.  Find  the  area  of  a  section  inclined  at 
an  angle  of  26°  37'  to  the  base,  one  of  the  sides  of  the  section 
being  parallel  to  the  base,  and  at  a  distance  therefrom  equal  to 
two-thirds  of  the  altitude. 

38 .  The  sides  of  an  inscribed  quadrilateral  are  a,  h,  c,  d. 
What  is  the  radius  of  the  circumscribing  circle? 


EXERCISES.  13 

|>9.  If  a  triangle  be  inscribed  in  ABC,  its  perimeter  must 
lie  between  a-\-b  +  G  and  a  cos  A  -f  ^  cos  B  +  c  cos  C,  (  =  4R  sin  A 
sin  B  sin  C). 


CHAPTER  V. 


1.   Solve  the  spherical  triangle  ABC,  given: — 

(1)  a  =106°  17' 22";  ft  =  120°  08' 20";  A=    39°  52' ; 

(2)  A  =  129°  05'  28"  ;  B  =  140°  25' ;         C  =  110°  27' ; 

(3)  a=   70°  32';  b=   38°  25' ;         C=   51°  17' 24"; 
(4)A=49°37';         B=126°;  ft=110°17'; 
(5)   C=    50°  32';         B=  136°  15' 28"  ;  •  a=    68°  35' . 

2  .  To  solve  a  spherical  triangle,  given 

c,  B,  and  A  +  C,  or  c,  B,  and  a  +  c. 

3.  Establish  the  propositions  concerning  spherical  triangles 
usually  given  in  the  text-books  in  Geometry,  by  means  of  the 
formulse  of  chapter  V. 

4.  Prove  directly  from  a  figure  that 

sin  a :  sin  c : :  sin  A :  siaC. 

5.  Show  that  the  values  given  (47)  for  sin|^,  tan  ^a,  etc., 
are  real. 

6.  Show  that  sin  a :  sin  A : : 

sin  b  sin  ciV^l  -  cos'^a  -  cos'^b  -  cos'^c  +  2  cos  a  cos  b  Cos  c. 

7.  Sliow  that  the  sum  of  any  two  angles  of  a  spherical  tri- 
angle is  less  than  180°  + the  third.  AYhat  property  of  a  right 
triangle  is  suggested  hereby  ? 

8.  If  OX,  OY,  OZ,  be  mutually  at  right  angles,  and  any 
two  lines  be  drawn  through  O, 

(1)  The  sum  of  the  squares  of  the  cosines  of  the  angles  that 
each  makes  with  OX,  OY,  and  OZ,  respectively,  will  equal 
unity. 

(2)  The  cosine  of  the  angle  between  the  two  lines  will  equal 
the  sum  of  the  products  of  tlie  cosines  of  the  angles  made  by 
the  two  lines,  with  OX,  with  OY,  and  with  OZ,  respectively. 


14  EXERCISES. 

9.  If   CD  be  a  median  in  ABC, 

cos  AC  +  cos  CB  =  2  cos  |AB  cos  CD. 

10.  If  two  angles  of  a  spherical  triangle  be  equal  to  the 
opposite  sides  respectively,  the  remaining  angle  will  be  equal  or 
supplementary  to  the  third  side. 

11.  D  is  any  point  in  AB ;  show  that 

cos  AD  sin  BC  =  cos  AB  sin  DC  +  cos  AC  sin  DB. 

12.  If  arcs  be  drawn  from  the  vertices  of  any  spherical 
triangle  through  any  point,  the  product  of  the  sines  of  one  set 
of  alternate  segments  of  the  sides  will  equal  the  product  of  the 
sines  of  the  other  set ;  and  conversely. 

IB.  The  last  proposition  proves  the  concurrence  of  what 
sets  of  angle-transversals  of  a  spherical  triangle  ? 

1 4 .  If  an  arc  cuts  the  sides  of  ABC,  in  D,  E,  and  F,  re- 
spectively, what  relation  obtains  between  the  two  sets  of  alter- 
nate segments  ? 

15.  If  ^,  /,  m,  be  the  altitudes  of  a  spherical  triangle, 

sin  k  sin  a  =  sin  Z  sin  b  =  sin  m  sin  c ; 
also,  cos  m  =  esc  c  V  cos^a  +  cos^ft  -  2  cos  a  cos  b  cos  c. 

1 6.  San  Francisco  is  122°  25'  40."  TCd  W.,  37°  AT  22."55  N., 
and  Yokohama  is  139°  40'  27"  K,  35°  26'  50"  N.;  what  is  the 
shortest  sailing  distance  between  the  cities?  What  is  the 
shortest  distance  on  the  surface  of  the  earth  between  Mount 
Hamilton  and  Annapolis',  Md.?     (See  Ex.  22). 

17.  Show  that  in  any  right  spherical  triangle  (A  =  90°). 

sin^Ja  =  sin"-J6  cos'^-Jc  +  cos%6  sin^Jc. 

18.  If  ^,  /?,  y^  be  the  lengths  of  the  angle-bisectrices  of 
ABC, 

cot  a  cos  ^A  +  cot  f3  cos  JB  +  cot  y  cos  ^C  =  cot  a  +  cot  b  4-  cot  c. 

19.  Determine  the  arcual  radii  (polar  distances),  of  the 
circumscribed,  inscribed,  and  escribed  circles,  of  the  triangles 
formed  by  the  intersection  of  any  three  arcs  (of  great  circles), 
on  the  sphere. 


EXERCISES.  15 

20.  ABC  is  equilateral,  P  the  pole  of  its  circumscribing 
circle,  and  Q  any  point  on  the  sphere.     To  show  that 

cos  QA  +  cos  QB  +  cos  QC  =  3  cos  PA  cos  PQ. 
What  is  the  analogous  formula  for  the  plane  triangle  ? 

21.  The  angle  of  elevation  of  the  top  of  a  cliff,  A,  is 
23°  27'  12" ;  the  angle  of  elevation  of  a  cliff,  B,  is  10°  28'  53"  ; 
the  angle  subtended  at  the  point  of  observation  by  AB  is 
73°  47'  18".     What  is  the  horizontal  angle  between  A  and  B  ?  * 

22.  ABC  is  trirectangular ;  P  is  the  pole  of  its  circum- 
scribed circle,  and  Q  any  point  on  the  sphere.     Show  that 

(cos  AQ  +  cos  BQ  +  cos  CQ)"'=  3  cosTQ. 
2  B.  In  any  triangle,  ^(a  +  c)  and  J(A  +  C)  agree  in  species. 

24.  If  r>,  E,  F,  be  the  feet  of  the  altitudes  of  ABC, 
tan  BD .  tan  CE .  tan  AF  --=  tan  DC .  tan  E  A .  tan  FB. 

25.  Through  any  point,  in  the  side  of  a  spherical  triangle, 
to  draw  an  arc  (of  a  great  circle),  that  shall  cut  off  a  given 
fracti(>n  of  the  area. 

2  6 .  Three  circles  are  described  in  a  triangle,  each  angle  of 
which  is  120°,  so  that  each  circle  touches  the  two  others  and 
two  sides  of  the  triangle.     Find  the  arcual  radii. 

27.  To  find  the  angle  formed  by  two  adjacent  faces  of  a 
regular  polyedron. 

28.  To  find,  for  any  regular  polyedron,  the  radii  of  the  in- 
scribed and  circumscribed  spheres. 

29.  To  find,  for  any  tetraedron  the  radii  of  the -circum- 
scribed, the  inscribed,  and  the  escribed,  spheres. 

30.  To  express  the  volume  of  any  tetraedron,  (and  hence 
of  any  parallelopiped),  in  terms  of  the  edges. 

31.  If,  wath  the  vertices  of  any  parallelopiped  as  centres, 
equal  spheres  be  described,  the  sum  of  the  intercepted  portions 
of  the  parallelopiped  will  equal  the  volume  of  any  one  of  the 
spheres. 

*The  solution  of  examples  of  this  character  is  called  "reducing  angles  to  the 
horizon." 


CHAPTER  VI. 

1.  Show  that : — 

(1)  cosh^cc  -  cosh'-i/  =  sinh'^a?  -  sinh'''2/=  sinh  (x  +  y)  sinh  {x  -  y) 

(2)  cosh^^  4-  sinh^o;  =  cosh^a;  -f  sinh^y  =  ^^sh  (x  +  y)  cosh  (x  -  y) ; 

^^,         ,  X         sinha?  .  .  i  ^*     r^       , 

(d)  tanh  ^=-  :; i — =  coth  a;  -  cscn  x^  coth  r  -  2  csch  a; ; 

^  '  2     1+  cosh  X  2 

1  +  tanh  X  tanh  y 

(4)  cosh(a;  +  y)       •= -. ^; 

^  '  ^       ^'  sech  re- sech y 

(5)  cosh  (n  +  2)^    =  2  cosh  k  cosh  (?i  +  1)A; — cosh  nk. 

What  are  the  corresponding  formulse  for  circular  functions? 

2.  Prove  that 

\  a  (p  +  qx^)  y  p{a  +  hx') 

What  is  the  corresponding  formulse  for  inverse  hyperbolic 
functions  ? 

3.  Show  that 

"/  (a  + 1)  cos  J-  a?  +  V{ct  -  h)  sin  J  x 

In ———-—-  = 

V  {a  +  ^)  cos  -J  a;  -  V{a  -  h)  sin  ^  x 

2  tanh~^  \     ,         tan  |-  x 
(Va  +  6 

Find  corresponding  logarithmic  equivalents  for 

tan~^  -!     tan  \x\- .  and  tanh~^  -^      .  tanh  -^  x 

(l^a  +  6  y  (l/a+6 

4.  Show  that : — 

( 1 )  cosh  (In  ?/)  +  sinh  (1  n  ?/)  =  2/ ; 

(2)  tanh  (In  y)^  {if  -  1)  +  {f  +  1).     Thence  show  that 

hi2=6.9;jl4718  +  . 

(3)co^.„^'[±|)=^;tanh(.n^[±i)=, 

(4)  If  f=-  1w  -  1,  tanh  (In  t/)  =  l-w-\ 

(5)  cosh-^(sec  x)  =  tanh-^(sin  x)  =  2  tanh-^(tan  \  x) 


(16) 


,     1  +  tan  -A-  a;     -  ,  /  -  \ 

=  ln  - — - — f— Intan^    13  +  ^    • 
1-tan^^  "\  2         / 


EXERCISES.  17 

(6)  If  tan-Jcc     =2,  "^ 

sinli  a:=  .j ;  cosh  x  = . 

1-2^  1  -  z^ 

(7)  OO.I.--  f-+^-  =  sinh--    laV  +  -2abx  +  ac 

v¥  -ac  \  6-  -  ac 

(8)  In tan  ^    . 

a — a?  a 

(9)ln^:±^^^=tanh-^. 

a;-^_^-/2  +  l  1  +  cc^ 

5.  Find  the  hyperbolic  functions  of  J-i,  ^-i,  and  y\7~i. 

6.  Calculate  directly  sinh  1  and  cosh  1 ;  thence  show  that 

e  =  2.71828182845904523536  + . 
Show  also  that 

sin  1  =  0.841470984807896  +  ; 
cos  1  =  0.540302305868039  +  . 

7.  Apply  De  Moivre's  theorem  to  hyperbolic  functions  so  as 
to  deduce,  for  the  powers  of  sinh  k  and  cosh  k  formulae  analo- 
gous to  those  given,  for  circular  functions,  on  pages  4,  5. 

8.  Show  that  :— 
\nsmhk=^k-\n2-e-''-ie-''-ie-'^-ie-'^-..  .  .  .; 
lncoshk  =  k-\n2  +  e-''-ie-'^  +  le-^-ie-''  + 

Thence  derive  the  development  of  In  tanh  k. 

9.  Show  that 

cosh  (£c  +  ^)  =  cosh  k  +  sinh  kj  +  cosh  k—  +  sinh.^-  .57  + .... ; 

sinh  (a;  +  ^)  =  sinh  k  +  cosh  h~-\- sinh  ^ t^  +  cosh X;—  + 

What  are  the  corresponding  developments  for  the  circular 
functions  ? 

10.  Prove  the  truth  of  the  following  equations : — 
cosh  2aj  =      2  cosh-oj  -      1, 

cosh  3ic  =     4  cosh^o;  -      3  cosh  x, 

cosh  4:c  =      8  cosh%  -      8  cosh^a;  +      1 , 

cosh  5^  =    16  cosh^rc  —   20  cosh^.?;  +      5  cosh  x. 


18  EXERCISES. 

cosh6ir=    32  cosher-    48cosh*a;+    IBcosh^a;-    1, 
cosh  7x  =    64  cosh'x  -  1 1 2  cosh^ic  +    56  cosh^a?  -    7  cosh  x, 
cosh  So;  =  128  cosh'o;  -  256  cosh'^  + 1 60  cosh*a;  -  32  cosh=^:c  +  1. 
Establish  the  corresponding  formulae  for  the  hyperbolic  sine, 
and  tangents.     Express  also  sinh^o:,  sinh^ic,  cosh^:c,  cosli^a;,  etc., 
in  terms  of  functions  of  x.     Compare  Ex.  4-,  page  5. 

11.  If  ?i  =  2r,  .T  =  a  +  ftcot<9, 

(A  +  I'B) {x  -  a  +  hiy  +  (A  -  iB)(a;  -  g  -  6i)» _ 
\{x-ay  +  b'\^'  ~ 

A  cos  n^' -  B  sin  ?i^  , 

\ix-ay  +  hY 

12.  Show  that,  as 

In  (cos  k  +  i  sin  k)  ^  ki, 

the  logarithm  of  unity  has  an  infinite  number  of  values,  of 
which  all  but  one  are  imaginary.  Thence  show  that  any 
number  has  an  infinite  number  of  logarithms  for  any  given 
base. 

1 3.  Show  that 


."-(.-;)(.-,4).... 


This  may  be  shown  l)y  proving,  first,  that  the  equation 
sin  ^  =  0  can  have  no  imaginary  roots,  and  then  showing  that 
the  series  (see  page  49)  for  sin  6  is  exactly  divisible  by  6±n-, 
n  being  any  integer.  Thus  sin  8  is  shown  to  be  a  product  of 
the  form 

a^(^+r)(^-r)(/9+2rr)(^-2ir)(^+3T:)(<9-3-) 

in  which  a  cannot  contain  ^,  since  the  equation  sin^  =  0  has 
no  imaginary  roots.  Finally  it  may  be  shown  that  a  =  l, 
whence  we  may  factor  sin  6  as  shown,  cos  6  may,  of  course, 
be  treated  similarly. 

From  these  expressions  for  sin  ^  and  cos  ^  we  may  deduce 
interesting  forms  of  algebraic  statement  for  the  other  expo- 
nential functions. 

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